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

22.

Q5

11.
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.

Q6

1e\dfrac{1}{e}.
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}.

Q7

(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.

Q8

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

Q9

eixe^{ix}.
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.

Q10

ex+e2\sqrt{e}x+\dfrac{\sqrt{e}}{2}.
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}.

Q11

ee.
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.

Q12


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

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

@Zakir Husain

Jeff Giff - 5 months, 3 weeks ago

Log in to reply

@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

Log in to reply

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

Log in to reply

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

Zakir Husain - 5 months, 2 weeks ago

Log in to reply

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

Jeff Giff - 5 months, 2 weeks ago

Log in to reply

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

Log in to reply

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

Jeff Giff - 5 months, 2 weeks ago

Log in to reply

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

Log in to reply

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

Jeff Giff - 5 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...