MA 1505 Tutorial 3: Taylor Series

The Taylor Series of f(x) at the point x_{0} is

f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}.

e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}

\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}

\sin x= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}

\cos x =\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}

\frac{1}{1-x} =\sum_{n=0}^{\infty} x^{n}

Question 1. Let S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)} . Calculate the value of S.

Solution.

Method (i).

S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)}

= \sum_{n=0}^{\infty} \frac{n+1}{(n+2)!}

= \sum_{n=0}^{\infty} \frac{(n+2)-1}{(n+2)!}

= \sum_{n=0}^{\infty} (\frac{1}{(n+1)!}-\frac{1}{(n+2)!})

= 1

Method (ii). Integrate the Taylor series of xe^{x} to show that S=1.

The Taylor series of x e^{x} is \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} . Take the integration of the function on the interval [0,1], we get

\int_{0}^{1} xe^{x} dx

=\int_{0}^{1} \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \frac{1}{n!(n+2)}=S .

The left hand side equals to 1 from integration by parts.

Method (iii). Differentiate the Taylor series of (e^{x}-1)/x.

The Taylor series of f(x)= (e^{x}-1)/x is \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} . Differentiate f(x) and get f^{'}(x)= \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!n} . Moreover, f^{'}(x)= \frac{e^{x}x-(e^{x}-1)}{x^{2}} and f^{'}(1)=1=S.

Method (iv). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)). This implies f(0)=0. Assume

g(x)=\int_{0}^{x}f(t)dt= \sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^{n}}{n!(n+2)}dt = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+2)!} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = \frac{1}{x}(e^{x}-1-x).

Since f(x)=g^{'}(x), we get f(x) = x^{-1}(e^{x}-1)-x^{-2}(e^{x}-1-x). That means f(1)=1.

Method (v). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)).

f(x)= \sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} - \frac{x^{n}}{(n+2)!} = x^{-1}\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} - x^{-2}\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = x^{-1} (e^{x}-1) - x^{-2}(e^{x}-1-x). Therefore, f(1)=1.

Remark. There is a similar problem: calculate \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!(n+2)}. Answer is 1-2e^{-1}.

Question 2. Let n be a positive integer. Prove that

\frac{1}{2} \int_{0}^{1} t^{n-1}(1-t)^{2} dt= \frac{1}{n(n+1)(n+2)}

and calculate the value of the summation

S=\frac{1}{1\cdot 2 \cdot 3} + \frac{1}{3\cdot 4 \cdot 5} + \frac{1}{5\cdot 6\cdot 7} + \frac{1}{7\cdot 8 \cdot 9}+.... .

Solution. 

\frac{1}{2}\int_{0}^{1} t^{n-1}(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} (t^{n+1}-2t^{n}+t^{n-1}) dt

= \frac{1}{2} (\frac{1}{n+2}-\frac{2}{n+1}+\frac{1}{n})

= \frac{1}{n(n+1)(n+2)} .

To calculate the value of S, there are two methods.

Method (i). The summation of S, n is only taken odd numbers. From the first step, we know the summation

S=\frac{1}{2} \int_{0}^{1} (1+t^{2}+t^{4}+t^{6}+...)(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} \frac{1}{1-t^{2}} (1-t)^{2} dt

= \frac{1}{2} \int_{0}^{1} \frac{1-t}{1+t} dt

= \frac{1}{2} \int_{0}^{1} (\frac{2}{1+t}-1)dt

= \frac{1}{2}( 2\ln(1+t)-t)_{t=0}^{t=1}

= \ln 2 -\frac{1}{2} .

Method (ii).

Since \frac{1}{n(n+1)(n+2)}= \frac{1}{2}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}) ,

S=\sum_{ odd} \frac{1}{n(n+1)(n+2)}

= \frac{1}{2} \sum_{odd} ( \frac{1}{n}- \frac{2}{n+1}+\frac{1}{n+2})

= \frac{1}{2} ( \frac{1}{1}-\frac{2}{2}+\frac{1}{3}+ \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+ \frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9}+...)

= \frac{1}{2} ( \frac{1}{1} + 2( -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...))

= \frac{1}{2} ( 1 + 2 (\ln 2-1))

= \ln 2 -\frac{1}{2} .

Here we use the Taylor series of \ln(1+x)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n} and \ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....

Question 3. Assume \zeta(k)=1+\frac{1}{2^{k}} + \frac{1}{3^{k}} + ... = \sum_{m=1}^{\infty} \frac{1}{m^{k}}.

Prove

\sum_{k=2}^{\infty} (\zeta(k)-1)=1.

\sum_{k=1}^{\infty} (\zeta(2k)-1)=3/4.

Proof.

\sum_{k=2}^{\infty} (\zeta(k)-1)

= \sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \frac{1}{(m-1)m}

= \sum_{m=2}^{\infty} ( \frac{1}{m-1} - \frac{1}{m})

= 1.

\sum_{k=1}^{\infty} ( \zeta(2k)-1)

= \sum_{k=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \frac{1}{m^{2}-1}

= \sum_{m=2}^{\infty} \frac{1}{2} ( \frac{1}{m-1}-\frac{1}{m+1})

= \frac{1}{2}(1+\frac{1}{2})

= \frac{3}{4}.

Question 4. Calculate the summation S= \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}.

Solution. 

S=\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}

=\frac{1}{4} \sum_{k=1}^{\infty} (-1)^{k} ( \frac{1}{2k-1} +\frac{1}{2k+1})

= \frac{1}{4} \sum_{k=1}^{\infty} ( \frac{(-1)^{k}}{2k-1} - \frac{(-1)^{k+1}}{2k+1})

= \frac{1}{4} \cdot \frac{-1}{2-1} = - \frac{1}{4}.

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