Increasing,Decreasing

Can exsinx e^x \sin x repeats its value other than 00 for atleast one time x>2\forall x > 2?

Note by Akash Shukla
3 years, 4 months 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

Actually,every value repeats infinite times.If you look at the graph,which is this,you'll notice that every value of the function repeats infinitely.Let's try to prove this mathematically.Let f(x)=exsinxf(x)={ e }^{ x }\sin { x } and lets choose 3=x13=x_1 for an example, we need to find another value x2x_2 such that f(3)=f(x2)f(3)=f(x_2) with x2>2x_2>2 as specified by you and of course,3x23\neq x_2.So we havee3sin3=ex2sinx2{ e }^{ 3 }\sin { 3 } ={ e }^{ { x }_{ 2 } }\sin { { x }_{ 2 } }

taking the natural logarithm on both sides,3+lnsin3=x2+lnsinx23+\ln { \sin { 3 } } ={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }

approximating sin3\sin{3} and plugging it in,

1.041=x2+lnsinx21.041={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }

the function on the RHS\text{RHS} repeats values too(we can check this graphically or mathematically,but mathematically proving it will take a while)

so there's an infinite number of solutions for x2x_2.We can do this for any value of x1x_1 and still end up with infinitely many values of x2x_2

Hamza A - 3 years, 4 months ago

Log in to reply

Thank you. After posting this question, I got my answer,not completely. Firstly I observed that exe^x increases for all xx and sin(x)sin(x) increases and decreases. So there will be such values such that function will repeat its value. But I found some problems with this. Lets say, for x=3x=3, f(x)=e3sin(3)f(x) = e^3*sin(3). 3π3≈π, and so f(x)f(x) will be very less. Now for x=2kπ+lx = 2k\pi+l function will repeat its value, ll depends on power of ee.If power of ee is larger, value of ll will be smaller. Now for repetition, e2π+lsin(l)=e3sin(3)e^{2\pi+l}*sin(l) = e^3*sin(3), l<π3l< \pi-3. But the problem is this only, is this condition be satisfied? As it is a function with exponents, so it is not necessary that it will satisfy the above condition, or I may be wrong.

Akash Shukla - 3 years, 4 months ago

Log in to reply

Sorry,but i'm not understanding what you're saying in the third line,can you please elaborate?

Hamza A - 3 years, 4 months ago

Log in to reply

@Hamza A If x=3x= 3 which is π≈\pi , then f(x)f(x) will be very less. Let this value be f(3)=Af(3) = A. Now I want to get AA for some higher value than 33. For this we have to take the value of xx greater than 2kπ2k\pi or less (2k+1)π (2k+1)\pi, as sinsin will be positive for such values. Let for x=2π+lx=2\pi+l ,f(x)f(x) repeats its value as that of f(3)f(3). Here ll must be such that, it should satisfy, e(2π+l)sin(l)=e(3)sin(3)e^{(2\pi+l)}*sin(l) = e^{(3)}*sin(3). Here value of xx is related with both the exe^x and sin(x)sin(x). So it will be very difficult to have same values as f(3)f(3), for different values of xx. As exe^x increases in a different way and sin(x)sin(x) changes in different way. But it is possible to have e(2π+l)sin(t)=e(3)sin(3)e^{(2\pi+l)}*sin(t) = e^{(3)}*sin(3), where t is nearer to ll or different than ll

Akash Shukla - 3 years, 4 months ago

Log in to reply

@Akash Shukla Ok,although i don't think a nice closed form for ll exists(if there is one),it's still there.You can approximate it by your method at the end by making another variable tt which is very near to ll,but to claim that it is not possible to have such ll just because the equation is very hard to solve exactly isn't a valid conclusion.

Hamza A - 3 years, 4 months ago

Log in to reply

@Hamza A 'Very hard' I mean to say that there will be hardly few values of ll to repeat the value, because it has to satisfy two different functions.

Akash Shukla - 3 years, 4 months ago

Log in to reply

@Akash Shukla how will it need to satisfy 2?it only needs to satisfy f(3)=f(2π+l)f(3)=f(2\pi+l)

Hamza A - 3 years, 4 months ago

Log in to reply

@Hamza A two different functions :- exe^x and sin(x)sin(x)

Akash Shukla - 3 years, 4 months ago

Log in to reply

@Akash Shukla but the function is them both multiplied,so it is technically just one function

Hamza A - 3 years, 4 months ago

Log in to reply

@Hamza A I think either you are not getting me or I'm not writing it properly. But anyway,now I will write it elaborately.

Now I will give you my calculative part done by calculator. I have first found the value of e3sin(3)e^3*sin(3) which comes to be 1.051195..1.051195... Now I have taken l=0.01l = 0.01, and found the value of e2π+0.01sin(0.11355..)=f(3)e^{2\pi+0.01}*sin(0.11355..) = f(3). So ll cannot be 0.010.01. Now here what I see is we have to increase the value of ll. But here exe^x function rapidly increase but sin(x) won't. So we couldn't get such value of ll to repeat the function.

Akash Shukla - 3 years, 4 months ago

Log in to reply

@Akash Shukla The function is continuous, so you can use the Intermediate Value Theorem.

Let f(x)=exsinx f(x) = e^x \sin x . Then f(2π)=0 f(2\pi ) = 0 and f(2π+π2)=e2π+π2>e6>1.051195=f(3) f(2\pi + \frac{\pi}{2} ) = e^{2 \pi + \frac{\pi}{2}} > e^6 > 1.051195 = f(3) .

Thus by IVT, we can conclude there exists an x0 x_0 between 2π 2\pi and 2π+π2 2\pi + \frac{\pi}{2} such that f(x0)=f(3) f(x_0) = f(3) .

Using the above format, you can find infinite values of x0 x_0 between 2nπ 2n\pi and 2nπ+π2 2n\pi + \frac{\pi}{2} as long as e2nπ+π2>f(3) e^{2n\pi + \frac{\pi}{2}} > f(3)

Siddhartha Srivastava - 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...