Increasing,Decreasing

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

#Calculus

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Comments

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 $x_2$

Hamza A - 5 years 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.

Akash Shukla - 5 years ago

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

Hamza A - 5 years ago

@Hamza A – 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 $l$

Akash Shukla - 5 years ago

@Akash Shukla – 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.

Hamza A - 5 years ago

@Hamza A – '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.

Akash Shukla - 5 years ago

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

Hamza A - 5 years ago

@Hamza A – two different functions :- $e^x$ and $sin(x)$

Akash Shukla - 5 years ago

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

Hamza A - 5 years ago

@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 $e^3*sin(3)$ 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.

Akash Shukla - 5 years ago

@Akash Shukla – 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 $e^{2n\pi + \frac{\pi}{2}} > f(3)$

Siddhartha Srivastava - 5 years ago

Math	Appears 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}`	$\boxed{123}$