# Happy e day! -solutions page

Happy e day!
Scoreboard: 1 attempt —— average points 9/12

## Q1

Euler’s number (see this)

## Q2

an irrational number

## Q3

False.
P.S. The function is nonzero.

## Q4

$2$.

## Q5

$1$.
$\displaystyle e=\lim _{x\to \infty} (1+\frac{1}{x})^x$. Therefore $\lim _{x\to \infty} \frac{1}{e} (1+x^{-1})^x =\frac{1}{e} \cdot e=1.$

## Q6

$\dfrac{1}{e}$.
Note that using Taylor series, $\displaystyle e^x =\sum ^{\infty} _{n=0} \frac{x^n}{n!}$. Therefore the sum is $e^{-1}=\frac{1}{e}$.

## Q7

$(2k+1)i\pi$,$k\in \mathbb{Z}$.
Euler’s formula states that $e^{ix}=\cos x +i\sin x$, where $i=\sqrt{-1}$. It is also commonly known that $e^{i\pi}=-1.$ Note that sine and cosine functions have a period of $2\pi$, so we have to generalise the solution.

## Q8

$e^x$.
This is the derivative of $e^x$, which is $e^x$.

## Q9

$e^{ix}$.
Again using Taylor series, $\cos x=\sum ^{\infty} _{n=0} \frac{(-1)^n x^{2n}}{(2n)!};$ $\sin x=\sum ^{\infty} _{n=0} \frac{(-1)^n x^{2n+1}}{(2n+1)!}.$ Therefore the sum=$\cos x+i\sin x=e^{ix}$ using Euler’s formula.

## Q10

$\sqrt{e}x+\dfrac{\sqrt{e}}{2}$.
The tangent of $y=e^x$ at $x=0.5$ has slope $\{ e^x \} ’=e^x=e^{0.5}=\sqrt{e}$. Let $f(x)=\sqrt{e} x+k$. $f(0.5)=e^{0.5}$ as well, solving to $k=\dfrac{\sqrt{e}}{2}$. So $f(x)=\sqrt{e} x+\dfrac{\sqrt{e}}{2}$.

## Q11

$e$.
Rearrange the first equation to get:
$b_{n+1}=\color{#3D99F6}b_n \color{#333333} (\dfrac{1}{a} -n)=\color{#3D99F6}b_{n-1}[\dfrac{1}{a}-(n-1)]\color{#333333} (\dfrac{1}{a} -n)=(\dfrac{1}{a}-n)\color{#3D99F6}[\dfrac{1}{a}-(n-1)]...(\dfrac{1}{a}-1)\dfrac{1}{a}\color{#333333}= \displaystyle \prod ^n _{k=0} (\dfrac{1}{a} -k).$
$\displaystyle \therefore b_{\color{#D61F06} n}=\prod ^{\color{#D61F06} n-1} _{k=0} (\dfrac{1}{a} -k)$.
\begin{aligned} \therefore \color{#D61F06} b_n \color{#333333} a^n &=a^n \color{#D61F06} \prod ^{n-1} _{k=0} (\dfrac{1}{a} -k)\\ &\color{#333333} =\prod ^{n-1} _{k=0} a(\frac{1}{a}-k) ~~~\text{Note that the} ~ a^n ~\text{, moved into the product, is now} ~a.\\ &=\prod ^{n-1} _{k=0} (1-ak).\end{aligned} $\therefore \displaystyle b_n a^n= \lim _{a\to 0} \prod ^{n-1} _{k=0} (1-ak)=\prod ^{n-1} _{k=0} \lim _{a\to 0} (1-ak)=\prod ^{n-1} _{k=0} 1=1^n =1.$
$\therefore \displaystyle \lim _{a\to 0} \frac{b_n a^n}{n!} =\frac{1}{n!}$.
Therefore the sum is equal to $\displaystyle \sum ^{\infty} _{n=0} \frac{1}{n!} =e$.

## Q12

To find $x(t)$, we can start from finding $z(t)$ that is designed for both real and complex numbers through trial. This way, we can generalise the formula as every complex formula(with one variable) can be rewritten in the form $Ae^{(ax+b)i}+c$, where $x$ is the only variable.

So say we let $z=z(t)=Ae^{(mt+n)i}+k,~z:\mathbb{C} \to \mathbb{C}$.
Then $\frac{d^2z}{dt^2}=-m^2Ae^{(mt+n)i}.$ _ STILL AT WORK _ Note by Jeff Giff
4 months, 2 weeks ago

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## Comments

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- 4 months, 2 weeks ago

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@Zakir Husain sir I am loading the solutions but I have no idea how to solve Q12. Can you help me?

- 4 months, 1 week ago

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Start by substituting $Ae^{i\alpha t + i\beta}+\gamma$ as a trial function.

Using the equation you can find values of $\alpha\space and \space\gamma$ .

Then take the real part of the trial function as $x(t)$, it will work.

- 4 months, 1 week ago

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@Jeff Giff - Answer of Q7. that you marked is actually wrong :(

- 4 months, 1 week ago

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What is the correct answer then? Is it not $i\pi$ or $\pi i$?

- 4 months, 1 week ago

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There are infinitely many solutions because if $x_1$ is a solution then $x_1+2i\pi$ will also be a solution.

- 4 months, 1 week ago

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Oh I see. Just checked out Math also fails series 2 to make sure.

- 4 months, 1 week ago

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In Q3, you didn't specify that the polynomial had to be nonzero. So it's technically true, but I can understand that you chose to put the answer as False.

- 4 months, 1 week ago

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Precisely. @Zakir Husain sir will you change the statement of the problem?

- 4 months, 1 week ago

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