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2 years ago

Evaluating expressions: 1. x + 9 when x = 3

Mathematics

1 answer:

balu736 [363]2 years ago

4 0

⇨ The numeric value of this expression = 12.

To solve this expression, just <u>replace '</u><u>x</u><u>'</u> by the <em>indicated value</em>, being the number 3, and <em>perform</em> the <em>indicated </em>math <em>operations</em>.

${\large \displaystyle \sf {1 \cdot x+9 }}$

${\large \displaystyle \sf {x=3 }}$

${\large \displaystyle \sf {1 \cdot 3+9 }}$

${\large \displaystyle \sf { 3+9 }}$

${\orange{\boxed{\boxed{\pink {\large \displaystyle \sf { \ 12 \ }}}}}}\\\\$

Therefore, we can conclude that, the correct value of this expression = 12.

${\orange{\boxed{\boxed{\pink {\large \displaystyle \sf { \ 12 \ }}}}}}\\\\$

<h3><em>Hope this helps! ❤️</em></h3>

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Solve using Fourier series.

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With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
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You should end up with

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$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

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I think is
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