Evaluate the series

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots $$

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots $$

miniliv4
2022-10-02
Answered

Evaluate the series

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots $$

$$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots $$

You can still ask an expert for help

beshrewd6g

Answered 2022-10-03
Author has **12** answers

One approach is as follows:

Let $f(t)$ denote a triangle wave with the precise form described here (we'll take L=1). As you can see in the link, f can be expanded into its Fourier series as

$$f(t)=\frac{8}{{\pi}^{2}}\sum _{n=1}^{\mathrm{\infty}}\frac{(-1{)}^{n-1}}{(2n-1{)}^{2}}\mathrm{sin}((2n-1)\pi t)$$

Now, just plug $t=\frac{1}{2}$ into both sides of the equation above.

Let $f(t)$ denote a triangle wave with the precise form described here (we'll take L=1). As you can see in the link, f can be expanded into its Fourier series as

$$f(t)=\frac{8}{{\pi}^{2}}\sum _{n=1}^{\mathrm{\infty}}\frac{(-1{)}^{n-1}}{(2n-1{)}^{2}}\mathrm{sin}((2n-1)\pi t)$$

Now, just plug $t=\frac{1}{2}$ into both sides of the equation above.

Denisse Fitzpatrick

Answered 2022-10-04
Author has **3** answers

Hint: On way to evaluate your series is using Parseval theorem. At first you need to find the fourier series of a function like $f(x)=x$ in $[-\pi ,\pi ]$ and then applying Parseval theorem.

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Given the initial problem:

$$\ddot{x}+4x=f(t),x(t=0)=3,\dot{x}(t=0)=-1$$

So I started:

$${s}^{2}(X(s)-sx(0)-\dot{x}(0)+4X(s)=\mathcal{L}(f(t))$$

Now substitute the given values:

$${s}^{2}X(s)-3s-(-1)+4X(s)=\mathcal{L}(f(t))$$

Rearranging:

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The answer give: The Laplace transform is of the form:

$$X(t)=A\mathrm{cos}2t+B\mathrm{sin}2t+\frac{1}{2}{\int}_{0}^{t}f(\tau )\mathrm{sin}2(t-\tau )d\tau $$

Is there anybody that can help me to get the given form?

$$\ddot{x}+4x=f(t),x(t=0)=3,\dot{x}(t=0)=-1$$

So I started:

$${s}^{2}(X(s)-sx(0)-\dot{x}(0)+4X(s)=\mathcal{L}(f(t))$$

Now substitute the given values:

$${s}^{2}X(s)-3s-(-1)+4X(s)=\mathcal{L}(f(t))$$

Rearranging:

$$X(s)({s}^{2}+4)-3s+1=\mathcal{L}(f(t))$$

The answer give: The Laplace transform is of the form:

$$X(t)=A\mathrm{cos}2t+B\mathrm{sin}2t+\frac{1}{2}{\int}_{0}^{t}f(\tau )\mathrm{sin}2(t-\tau )d\tau $$

Is there anybody that can help me to get the given form?

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