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# System prepended metadata

title: Homework 8
tags: [微分方程特論]

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# Homework 8
###### tags: `微分方程特論`
by B05202054 何信佑
site: https://hackmd.io/@ulynx/topdehw8
issued on 20 May, 2020.
## Section 8.1
### Problem 30
Let $f(x)$ be a piecewise function with period $P$. **(a)** Suppose that $0\leq a<P$. Substitute $u=t-P$ to show that $$\int^{a+P}_P f(t)\text{d}t=\int^a_0 f(t)\text{d}t.\tag{i}$$ Conclude that $$\int^{a+P}_a f(t)\text{d}t=\int^P_0 f(t)\text{d}t.\tag{ii}$$ **(b)** Given $A$, choose $n$ so that $A=nP+a$ with $0\leq 0 <P$. Then substitude $v=t-np$ to show that $$\int^{A+P}_{A}f(t)\text{d}t=\int^{a+P}_{a}f(t)\text{d}t=\int^{P}_{0}f(t)\text{d}t.\tag{iii}$$

#### Solution of part (a)
Applying a change of variable $u=t-P$ (and thus $\text{d}u=\text{d}t$), we have $$\begin{align}\int^{a+P}_P f(t)\text{d}t&=\int^{a+P-P}_{P-P} f(u+P)\text{d}u=\int^{a}_{0} f(u+P)\text{d}u\\&=\int^a_0f(u)\text{d}u=\int^a_0f(t)\text{d}t,\end{align}$$ where we translated the integrand by $P$ because $f$ is a periodic funciton with period $P$, and then we replaced $u$ by $t$. 

From one of the basic rules of integration, we write $$\require{cancel}\int^{a+P}_0f(t)\text{d}t=\int^{P}_{0} f(t)\text{d}t+\int^{a+P}_{P} f(t)\text{d}t=\int^{a}_{0} f(t)\text{d}t+\int^{a+P}_{a} f(t)\text{d}t,$$ but Eqn. (i) allows us to do cancellation $$\int^{P}_{0} f(t)\text{d}t+\cancel{\int^{a+P}_{P} f(t)\text{d}t}=\cancel{\int^{a}_{0} f(t)\text{d}t}+\int^{a+P}_{a} f(t)\text{d}t$$ and lead us to conclude Eqn. (ii). $\blacksquare$

#### Solution of part (b)
Applying a change of variable $v=t-nP$ (and thus $\text{d}v=\text{d}t$), we have $$\begin{align}\int^{A+P}_{A}f(t)\text{d}t&=\int^{A+P+nP}_{A+nP}f(v+nP)\text{d}v=\int^{a+P}_{a}f(v+nP)\text{d}v\\&=\int^{a+P}_{a}f(v)\text{d}v=\int^{a+P}_{a}f(t)\text{d}t=\int^{P}_{0}f(t)\text{d}t,\end{align}$$ where the periodity of $f$ validates the third equality sign and Eqn. (ii) validates the last. $\blacksquare$

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## Section 8.2
### Problem 15
**(a)** Suppose that $f$ is a function of $2\pi$ with $f(t)=t^2$ for $0<t<2\pi$. Show that $$f(t)=\dfrac{4\pi^2}{3}+4\sum^\infty_{n=1}\dfrac{\cos nt}{n^2}-4\pi \sum^\infty_{n=1}\dfrac{\sin nt}{n}\tag{i}$$ and sketch the graph of $f$, indicating the value at each discontinuity. 

**(b)** Deduce the series summations in Eqs. (16) and (17) the Fourier series in part (a)
#### Solution of part (a)
If a piecewise continuous function $f(x)$ of ==period $2L$== is given by a single formula for ==$0<t<2L$==, then the **Fourier coefficients** are defined to be (Eqs. (10a), (10b) in the textbook) $$a_0=\frac{1}{L}\int^{2L}_{0}  f(t)\text{d}t,\tag{ii}$$ $$a_n=\frac{1}{L}\int^{2L}_{0}  f(t)\cos \dfrac{n\pi t}{L}\text{d}t,\quad n=1,2,3\ldots,\tag{iii}$$ and $$b_n=\frac{1}{L}\int^{2L}_{0} f(t)\sin \dfrac{n\pi t}{L}\text{d}t,\quad n=1,2,3\ldots.\tag{iv}$$

And then we evaluate the integrals in Eqs. (ii) to (iv) with the given $f(t)=t^2$ for ==$0<t<2\pi$== ($\because$ period = $2\pi$, $\therefore$ we put $L=\pi$ into those formulae.):
$\begin{align}a_0=\frac{1}{\pi}\int^{2\pi}_{0}  t^2\text{d}t=\dfrac{1}{\pi}\left[\dfrac{1}{3}t^3\right]^{2\pi}_{0} =\boxed{\dfrac{8\pi^2}{3}},\end{align}$
$\begin{align}a_n&=\frac{1}{\pi}\int^{2\pi}_{0}  t^2\cos nt\,\text{d}t=\frac{1}{\pi}\Bigg\{\cancelto{0}{\left[\dfrac{t^2}{n}\sin nt\right]^{2\pi}_{0}}-\int^{2\pi}_{0} \dfrac{2t}{n}\sin nt\,\text{d}t\Bigg\}\\&=-\frac{2}{n\pi}\int^{2\pi}_{0} t\sin nt\,\text{d}t=-\frac{2}{n\pi}\left\{\left[-\dfrac{t}{n}\cos nt\right]^{2\pi}_{0}-\int^{2\pi}_{0}\left[-\dfrac{1}{n}\cos nt\right]\text{d}t\right\}\\&=\dfrac{2}{n^2\pi}\Bigg\{(2\pi-0)-\cancelto{0}{\left[\dfrac{\sin nt}{n}\right]^{2\pi}_0}\Bigg\}=\boxed{\dfrac{4}{n^2}} \quad\text{for }n=1,2,3,\ldots,\end{align}$
$\begin{align}b_n&=\dfrac{1}{\pi}\int^{2\pi}_{0}t^2\sin nt\,\text{d}t=\dfrac{1}{\pi}\Bigg\{\left[-\dfrac{t^2}{n}\cos nt\right]^{2\pi}_{0}-\int^{2\pi}_{0}\left[-\dfrac{2t}{n}\cos nt\right]\text{d}t\Bigg\}\\&=-\dfrac{1}{n\pi}\left(4\pi^2-2\int^{2\pi}_0 t\cos nt\,\text{d}t\right)\\&=-\dfrac{4\pi}{n}+\dfrac{2}{n\pi}\Bigg\{\cancelto{0}{\left[\dfrac{t}{n}\sin nt\right]^{2\pi}_0}-\int^{2\pi}_0 \dfrac{1}{n}\sin nt\,\text{d}t\Bigg\}\\&=-\dfrac{4\pi}{n}-\dfrac{2}{n\pi}\cancelto{0}{\left[\dfrac{-\cos nt}{n^2}\right]^{2\pi}_0}=\boxed{-\dfrac{4\pi}{n}} \quad\text{for }n=1,2,3,\ldots.\end{align}$

Hence the **Fourier series** of $f$ is $$\begin{align}f(t)&\sim\frac{1}{2}a_0+\sum^\infty_{n=1}\left(a_n\cos nt+b_n\sin nt\right)\\&=\boxed{\dfrac{4\pi^2}{3}+\sum^\infty_{n=1}\left(\dfrac{4}{n^2}\cos n t-\dfrac{4\pi}{n}\sin n t\right)},\tag{v}\end{align}$$ which converges to $$\frac{f(t+)+f(t-)}{2}=\left\{\begin{array}{ll} 2\pi^2&\text{if }x=2k\pi~\text{with}~ k\in\mathbb{Z},\\f(t)&\text{otherwise},\end{array}\right.$$ because $f$ is periodic and piecewise smooth. The redefinition of domain with $$f(t)=\left\{\begin{array}{ll}2\pi^2&\text{if }x=2k\pi~\text{with}~ k\in\mathbb{Z},\\t^2&\text{otherwise},\end{array}\right.\tag{vi}$$ renders Series (v) convergent to $f(t)$ for all $t$. $\blacksquare$

![](https://i.imgur.com/0grdC9e.png =450x)
<font size=2>**Figure 1:** the graph of $f$</font>
\
![](https://i.imgur.com/z5nj2wV.png =450x)
<font size=2>**Figure 2:** the partial sum of the first $N$ terms of Series (iii) (from top to bottom: $N=1$, $N=5$, $N=11$, $N=99$)</font>


#### Solution to part (b)
Putting $t=0$ into Eqs. (v) and (vi) in part (a) and equating them, we obtain a famous series summation (vi): $$\begin{align}&2\pi^2=\dfrac{4\pi^2}{3}+\sum^\infty_{n=1}\Big(\dfrac{4}{n^2}\cancelto{1}{\cos 0}-\dfrac{4\pi}{n}\cancelto{0}{\sin 0}\Big)\\\implies&\boxed{\dfrac{\pi^2}{6}=\sum^\infty_{n=1}\dfrac{1}{n^2}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots}.\tag{vi}\end{align}$$

Likewise, putting $t=\pi$ into Eqs. (v) and (vi) in part (a) and equating them again, we obtain the other series summation (vii): $$\begin{align}&\pi^2=\dfrac{4\pi^2}{3}+\sum^\infty_{n=1}\Big(\dfrac{4}{n^2}\cos n\pi -\dfrac{4\pi}{n}\cancelto{0}{\sin n\pi }\Big)\\&\\\implies&\boxed{\dfrac{\pi^2}{12}=-\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots}.  \blacksquare\tag{vii}\end{align}$$

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### Problem 17
**(a)** Suppose that $f$ is a function of period $2$ with $f(t)=t$ for $0<t<2$. Show that $$f(t)=1-\dfrac{2}{\pi}\sum^\infty_{n=1}\dfrac{\sin n\pi t}{n}\tag{1}$$ and sketch the graph of $f$, inidcating the value at each discontinuity. 

**(b)** Substitute an appropriate value ot  $t$ to deduce **Leibniz's series** $$1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots=\dfrac{\pi}{4}\tag{ii}$$

#### Solution of part (a)
We calculate the Fourier coefficients of $f$ by Eqs. (ii) to (iv) in part (a) of Probelm 15, Section 8.2: (this time, use $L=1$)
$\displaystyle a_0=\int^2_0 t\,\text{d}t=\left[\dfrac{t^2}{2}\right]^2_0=\boxed{2},$
$\begin{align}a_n&=\int^2_0  t\cos n\pi t\,\text{d}t=\cancelto{0}{\left[\dfrac{t}{n\pi}\sin n\pi t\right]^2_0}-\int^2_0\dfrac{\sin n\pi t}{n\pi}\text{d}t\\&=-\left[-\dfrac{\cos n\pi t}{(n\pi)^2}\right]^2_0=\boxed{0}\qquad \text{for}~n=1,2,3,\ldots,\end{align}$
$\begin{align}b_n&=\int^2_0 t\sin n\pi t\,\text{d}t=\left[-\dfrac{t}{n\pi}\cos n\pi t\right]^2_0-\int^2_0\left(-\dfrac{\cos n\pi t}{n\pi}\right)\text{d}t\\&=-\dfrac{2}{n\pi}-\cancelto0{\left[-\dfrac{\sin n\pi t}{(n\pi)^2}\right]^2_0}=\boxed{-\dfrac{2}{n\pi}}\qquad \text{for}~n=1,2,3,\ldots\end{align}$

Because $f$ is periodic and piecewise smooth, its Fourier series $$\begin{align}f(t)&\sim\frac{1}{2}a_0+\sum^\infty_{n=1}\left(a_n\cos n\pi t+b_n\sin n\pi t\right)\\&=\boxed{1-\dfrac{2}{\pi}\sum^\infty_{n=1}\dfrac{\sin n\pi t}{n}}\tag{iii}\end{align}$$ converges to $$\frac{f(t+)+f(t-)}{2}=\left\{\begin{array}{ll} 1&\text{if }t=2k,\  k\in\mathbb{Z},\\f(t)&\text{otherwise},\end{array}\right.$$ where the discontinuities are $t=2k\ (k\in\mathbb{Z})$. $\blacksquare$

![](https://i.imgur.com/S5lZrkQ.png =450x)
<font size=2>**Figure 1:** the graph of $f$</font>
\
![](https://i.imgur.com/XLMrz7W.png =450x)
<font size=2>**Figure 2:** the partial sum of the first $N$ terms of Series (iii) (from top to bottom: $N=1$, $N=5$, $N=9$, $N=100$)</font>



#### Solution of part (b)
Putting into $t=1/2$ into Eqn. (iii) and equating with $f(t)=t$, we can deduce the so-called Leibniz’s series $$\begin{align}&\dfrac{1}{2}=1-\dfrac{2}{\pi}\sum^\infty_{n=1}\dfrac{\sin (n\pi/2)}{n}\\\implies&\boxed{\dfrac{\pi}{4}=\sum^\infty_{n=1}\dfrac{\sin (n\pi/2)}{n}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots}.\  \blacksquare\end{align}$$

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## Section 8.3
### Problem 3
Given the function $f(t)=1-t$ on the interval $0<t<2$, find the Fourier cosine and sine series of $f$ and sketch the graphs of the two extensions of $f$ to which these series converge.
#### Solution
$L=2$
1. We calculate the coefficients of **Fourier cosine series**: $$\begin{align}  a_0&=\dfrac{2}{L}\int^L_0 f(t)\,\text{d}t=\dfrac{2}{2}\int^2_0 (1-t)\text{d}t=\left[t-\dfrac{t^2}{2}\right]^2_0=0\end{align}$$ and $$\begin{align} a_n&=\frac{2}{L}\int^L_0  f(t)\cos \frac{n\pi t}{L} \text{d}t=\frac{2}{2}\int^2_0 \left[(1-t)\cos \frac{n\pi t}{2}\right]\text{d}t\\&=\int^2_0\cos \left(\frac{n\pi t}{2}\right)\text{d}t-\int^2_0\left(t\cos \frac{n\pi t}{2}\right)\text{d}t\\&=\left[\dfrac{\sin (n\pi t/2)}{n\pi/2}-\dfrac{t\sin (n\pi t/2)}{n\pi/2}+\dfrac{\cos (n\pi t/2)}{(n\pi/2)^2}\right]^2_0\\&=\dfrac{\cos (n\pi)-1}{(n\pi/2)^2}=\dfrac{(-1)^n-1}{(n\pi/2)^2}=\left\{\begin{array}{ll}-\dfrac{8}{n^2\pi^2}&\text{if}~~n=1,3,5,\ldots,\\0&\text{if}~~n=2,4,6,\ldots.\end{array}\right.\end{align}$$ (Only odd terms survive.)
    
    Thus the Fourier cosine series of $f$ is $$\begin{align}f(t)&\sim\frac{1}{2}a_0+\sum^\infty_{n=1}a_n\cos \frac{n\pi t}{L}\\&=\boxed{-\dfrac{8}{\pi^2}\sum^\infty_{k=1}\dfrac{\cos[(2k-1)\pi t/2]}{(2k-1)^2}}\tag{ii}\end{align}$$ which coverges to $$\frac{f(t+)+f(t-)}{2}=\left\{\begin{array}{ll} 1&\text{at}~~t=0,\\-1&\text{at}~~t=2,\\f(t)&\text{within}~~0<t<2,\end{array}\right.$$ or to the  **even period $2$ extension of $f$**, which is defined as $$\boxed{f_\text{E}(t)=\left\{\begin{array}{ll}f(t)&\text{if}~~0\leq t<2,\\f(-t)&\text{if}~~-2<t\leq0\end{array}\right.}\tag{iii}$$ and by $f_\text{E}(t+2)=f(t)$ for all $t$. 
    ![](https://i.imgur.com/HSZvpul.png =450x)
    <font size=2>**Figure 1:** the even period $2$ extension of $f$</font>



    
2. We calculate the coefficients of **Fourier sine series**: $$\begin{align} b_n&=\frac{2}{L}\int^L_0  f(t)\sin \frac{n\pi t}{L} \text{d}t=\frac{2}{2}\int^2_0 \left[(1-t)\sin \frac{n\pi t}{2}\right]\text{d}t\\&=\int^2_0\sin \left(\frac{n\pi t}{2}\right)\text{d}t-\int^2_0\left(t\sin \frac{n\pi t}{2}\right)\text{d}t\\&=\left[-\dfrac{\cos (n\pi t/2)}{n\pi/2}+\dfrac{t\cos (n\pi t/2)}{n\pi/2}-\dfrac{\sin (n\pi t/2)}{(n\pi/2)^2}\right]^2_0\\&=\dfrac{-[\cos (n\pi)-1]+2\cos (n\pi)}{n\pi/2}=\dfrac{(-1)^n+1}{n\pi/2}\\&=\left\{\begin{array}{ll}0&\text{if}~~n=1,3,5,\ldots,\\\dfrac{4}{n\pi}&\text{if}~~n=2,4,6,\ldots.\end{array}\right.\end{align}$$ (Only even terms survive.)
    
    Thus the Fourier sine series of $f$ is $$\begin{align}f(t)&\sim\sum^\infty_{n=1}b_n\sin \frac{n\pi t}{L}\\&=\boxed{\dfrac{2}{\pi}\sum^\infty_{k=1}\dfrac{\sin(k\pi t)}{k}}\tag{iii}\end{align}$$ which coverges to $$\frac{f(t+)+f(t-)}{2}=\left\{\begin{array}{ll} 0&\text{at}~~t=0,\\0&\text{at}~~t=2,\\f(t)&\text{within}~~0<t<2.\end{array}\right.$$ or to the  **odd period $2$ extension of $f$**, which is defined as $$\boxed{f_\text{O}(t)=\left\{\begin{array}{ll}f(t)&\text{if}~~0<t<2,\\0&\text{if}~~t=0\\-f(-t)&\text{if}~~-2<t<0\end{array}\right.}\tag{iv}$$ and by $f_\text{O}(t+2)=f(t)$ for all $t$. $\blacksquare$ 
    ![](https://i.imgur.com/cpIohB7.png =450x)
    <font size=2>**Figure 2:** the odd period $2$ extension of $f$</font>
    
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### Problem 19
Begin with the Fourier series $$t=2\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n}\sin nt,\qquad -\pi<t<\pi,\tag{i}$$ and integrate termise three times in succession to obtain the series $$\dfrac{1}{24}t^4=\dfrac{\pi^2t^2}{12}-2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^4}\cos nt+2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^4}.\tag{ii}$$
#### Solution
Integrate (termwise) Eqn. (i) for the first time: $$\begin{align} \dfrac{1}{2}t^2&=\int^t_0\left[ 2\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n}\sin n\tau\right]\text{d}\tau=2\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n}\int^t_0\sin n\tau\,\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n}\left[-\dfrac{\cos n\tau}{n}\right]^t_0=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}[\cos (nt)-1];\tag{iii}\end{align}$$ 

for the second time: $$\begin{align} \dfrac{1}{6}t^3&=\int^t_0\left\{2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}[\cos (n\tau)-1]\right\}\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\int^t_0[\cos (n\tau)-1]\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\left[\dfrac{\sin n\tau}{n}-\tau\right]^t_0=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\left[\dfrac{\sin n t}{n}-t\right];\tag{iv}\end{align}$$

for the third time: $$\begin{align} \dfrac{1}{24}t^4&=\int^t_0\left\{2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\left[\dfrac{\sin n \tau}{n}-\tau\right]\right\}\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\int^t_0\left[\dfrac{\sin n \tau}{n}-\tau\right]\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^n}{n^2}\left[-\dfrac{\cos n \tau}{n^2}-\dfrac{\tau^2}{2}\right]^t_0\text{d}\tau\\&=2\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n^2}\left[\dfrac{\cos (nt)-1}{n^2}+\dfrac{t^2}{2}\right]\\&=\sum^\infty_{n=1}\left[\dfrac{(-1)^{n}}{n^2}\right]t^2+2\sum^\infty_{n=1}\left[\dfrac{(-1)^{n+1}\cos(nt)-(-1)^{n+1}}{n^4}\right]\\&=\boxed{\dfrac{\pi^2t^2}{12}-2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\cos nt\right]+2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]}.\ \blacksquare\tag{v}\end{align}$$

### Problem 20
Substitute $t=\pi/2$ and $t=\pi$ in the series of Problem 19 to obtain the summations $$\sum^\infty_{n=1}\dfrac{1}{n^4}=\dfrac{\pi^4}{90},\qquad \sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n^4}=\dfrac{7\pi^4}{720}$$ and $$1+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\dfrac{1}{7^4}+\cdots=\dfrac{\pi^4}{96}.$$
#### Solution
Substituting $t=\pi/2$ into Eqn. (v) in Problem 19, Section 8.3, we find $$\begin{align}&\dfrac{1}{24}\left(\dfrac{\pi}{2}\right)^4=\dfrac{\pi^2}{12}\left(\dfrac{\pi}{2}\right)^2-2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\cos \dfrac{n\pi}{2}\right]+2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]\\\implies&\dfrac{7}{48}\left(\dfrac{\pi}{2}\right)^4=\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\cos \dfrac{n\pi}{2}\right]-\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]\\&\hspace{4.0em}=\left(-\dfrac{1}{2^4}+\dfrac{1}{4^4}-\dfrac{1}{6^4}+\cdots\right)-\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]\\&\hspace{4.0em}=\dfrac{1}{2^4}\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]-\sum^\infty_{n=1}\left[\dfrac{(-1)^{n}}{n^4}\right]=-\dfrac{15}{2^4}\sum^\infty_{n=1}\dfrac{(-1)^n}{n^4}\\\implies&\boxed{\dfrac{7\pi^4}{720}=\sum^\infty_{n=1}\dfrac{(-1)^{n+1}}{n^4}}.\tag{i}\end{align}$$ 

Substituting $t=\pi$ into the same equation, we find $$\begin{align}&\dfrac{\pi^4}{24}=\dfrac{\pi^4}{12}-2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\cos n\pi\right]+2\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]\\\implies&\dfrac{\pi^4}{48}=\sum^\infty_{n=1}\bigg[\dfrac{\cancelto{1}{(-1)^n(-1)^n}}{n^4}\bigg]-\underbrace{\sum^\infty_{n=1}\left[\dfrac{(-1)^n}{n^4}\right]}_{=\ -7\pi^4/720}\\\implies&\dfrac{\pi^4}{48}-\dfrac{7\pi^4}{720}=\boxed{\dfrac{\pi^4}{90}=\sum^\infty_{n=1}\dfrac{1}{n^4}}.\tag{ii}\end{align}$$

Substracting Eqn. (ii) from Eqn. (i), we obtain $$\begin{align}&\phantom{=}\sum^\infty_{n=1}\left[\dfrac{1}{n^4}\right]-\sum^\infty_{n=1}\left[\dfrac{(-1)^{n}}{n^4}\right]\\&=\left(\tfrac{1}{1^4}+\cancel{\tfrac{1}{2^4}}+\tfrac{1}{3^4}+\cancel{\tfrac{1}{4^4}}+\cdots\right)-\left(-\tfrac{1}{1^4}+\cancel{\tfrac{1}{2^4}}-\tfrac{1}{3^4}+\cancel{\tfrac{1}{4^4}}-\cdots\right)\\&=2\left(\dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\cdots\right)=\dfrac{\pi^4}{90}-\left(-\dfrac{7\pi^4}{720}\right)=\dfrac{\pi^4}{48}\\\implies&\boxed{\dfrac{\pi^4}{96}=\dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\cdots}.\ \blacksquare\end{align}$$



