×

# Increasing,Decreasing

Can $$e^x \sin x$$ repeats its value other than $$0$$ for atleast one time $$\forall x > 2$$?

Note by Akash Shukla
1 year, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $$f(x)={ e }^{ x }\sin { x }$$ and lets choose $$3=x_1$$ for an example, we need to find another value $$x_2$$ such that $$f(3)=f(x_2)$$ with $$x_2>2$$ as specified by you and of course,$$3\neq x_2$$.So we have${ e }^{ 3 }\sin { 3 } ={ e }^{ { x }_{ 2 } }\sin { { x }_{ 2 } }$

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

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

$1.041={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }$

the function on the $$\text{RHS}$$ 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 $$x_2$$.We can do this for any value of $$x_1$$ and still end up with infinitely many values of $$x_2$$

- 1 year, 5 months ago

Thank you. After posting this question, I got my answer,not completely. Firstly I observed that $$e^x$$ increases for all $$x$$ and $$sin(x)$$ 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 $$x=3$$, $$f(x) = e^3*sin(3)$$. $$3≈π$$, and so $$f(x)$$ will be very less. Now for $$x = 2k\pi+l$$ function will repeat its value, $$l$$ depends on power of $$e$$.If power of $$e$$ is larger, value of $$l$$ will be smaller. Now for repetition, $$e^{2\pi+l}*sin(l) = e^3*sin(3)$$, $$l< \pi-3$$. 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.

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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

- 1 year, 5 months ago

how will it need to satisfy 2?it only needs to satisfy $$f(3)=f(2\pi+l)$$

- 1 year, 5 months ago

two different functions :- $$e^x$$ and $$sin(x)$$

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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 $$e^3*sin(3)$$ which comes to be $$1.051195..$$. Now I have taken $$l = 0.01$$, and found the value of $$e^{2\pi+0.01}*sin(0.11355..) = f(3)$$. So $$l$$ cannot be $$0.01$$. Now here what I see is we have to increase the value of $$l$$. But here $$e^x$$ function rapidly increase but sin(x) won't. So we couldn't get such value of $$l$$ to repeat the function.

- 1 year, 5 months ago

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

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

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

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

- 1 year, 5 months ago