Happy e day! -solutions page

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


Euler’s number (see this)


an irrational number


P.S. The function is nonzero.




e=limx(1+1x)x\displaystyle e=\lim _{x\to \infty} (1+\frac{1}{x})^x. Therefore limx1e(1+x1)x=1ee=1.\lim _{x\to \infty} \frac{1}{e} (1+x^{-1})^x =\frac{1}{e} \cdot e=1.


Note that using Taylor series, ex=n=0xnn!\displaystyle e^x =\sum ^{\infty} _{n=0} \frac{x^n}{n!}. Therefore the sum is e1=1ee^{-1}=\frac{1}{e}.


(2k+1)iπ(2k+1)i\pi,kZk\in \mathbb{Z}.
Euler’s formula states that eix=cosx+isinxe^{ix}=\cos x +i\sin x, where i=1i=\sqrt{-1}. It is also commonly known that eiπ=1.e^{i\pi}=-1. Note that sine and cosine functions have a period of 2π2\pi, so we have to generalise the solution.


This is the derivative of exe^x, which is exe^x.


Again using Taylor series, cosx=n=0(1)nx2n(2n)!;\cos x=\sum ^{\infty} _{n=0} \frac{(-1)^n x^{2n}}{(2n)!}; sinx=n=0(1)nx2n+1(2n+1)!.\sin x=\sum ^{\infty} _{n=0} \frac{(-1)^n x^{2n+1}}{(2n+1)!}. Therefore the sum=cosx+isinx=eix\cos x+i\sin x=e^{ix} using Euler’s formula.


The tangent of y=exy=e^x at x=0.5x=0.5 has slope {ex}=ex=e0.5=e\{ e^x \} ’=e^x=e^{0.5}=\sqrt{e}. Let f(x)=ex+kf(x)=\sqrt{e} x+k. f(0.5)=e0.5f(0.5)=e^{0.5} as well, solving to k=e2k=\dfrac{\sqrt{e}}{2}. So f(x)=ex+e2f(x)=\sqrt{e} x+\dfrac{\sqrt{e}}{2}.


Rearrange the first equation to get:
bn+1=bn(1an)=bn1[1a(n1)](1an)=(1an)[1a(n1)]...(1a1)1a=k=0n(1ak).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).
bn=k=0n1(1ak)\displaystyle \therefore b_{\color{#D61F06} n}=\prod ^{\color{#D61F06} n-1} _{k=0} (\dfrac{1}{a} -k).
bnan=ank=0n1(1ak)=k=0n1a(1ak)   Note that the an , moved into the product, is now a.=k=0n1(1ak).\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} bnan=lima0k=0n1(1ak)=k=0n1lima0(1ak)=k=0n11=1n=1.\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.
lima0bnann!=1n!\therefore \displaystyle \lim _{a\to 0} \frac{b_n a^n}{n!} =\frac{1}{n!}.
Therefore the sum is equal to n=01n!=e\displaystyle \sum ^{\infty} _{n=0} \frac{1}{n!} =e.


To find x(t)x(t), we can start from finding z(t)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+cAe^{(ax+b)i}+c, where xx is the only variable.

So say we let z=z(t)=Ae(mt+n)i+k, z:CCz=z(t)=Ae^{(mt+n)i}+k,~z:\mathbb{C} \to \mathbb{C}.
Then d2zdt2=m2Ae(mt+n)i.\frac{d^2z}{dt^2}=-m^2Ae^{(mt+n)i}. _ STILL AT WORK _

Note by Jeff Giff
5 months, 3 weeks ago

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@Zakir Husain

Jeff Giff - 5 months, 3 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?

Jeff Giff - 5 months, 2 weeks ago

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Start by substituting Aeiαt+iβ+γAe^{i\alpha t + i\beta}+\gamma as a trial function.

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

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

Zakir Husain - 5 months, 2 weeks ago

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

Zakir Husain - 5 months, 2 weeks ago

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

Jeff Giff - 5 months, 2 weeks ago

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There are infinitely many solutions because if x1x_1 is a solution then x1+2iπx_1+2i\pi will also be a solution.

Zakir Husain - 5 months, 2 weeks ago

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

Jeff Giff - 5 months, 2 weeks 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.

Elijah L - 5 months, 2 weeks ago

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

Jeff Giff - 5 months, 2 weeks ago

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