Number of real solution of expression containing definite integral

Asked Mar 19 '23 at 13:21

Active Mar 27 '23 at 07:19

Viewed 167 times

Let $f:[a,b]\rightarrow \mathbb{R}$ be a continuous function and $$\displaystyle \int^{b}_af(x)\sin\left(\frac{x-a}{b-a}\pi\right)\,dx=\int^b_af(x)\cos\left(\frac{x-a}{b-a}\pi\right)\,dx=0.$$

Then number of solution of $f(x)=0$ in $(a,b)$ is

Let we assume $\displaystyle g(x)=\int^b_a f(x)\sin\bigg(\frac{x-a}{b-x}\pi\bigg)dx$

Then $\displaystyle g(a)=0$ and $g(b)=0$

Then here using rolles Theorem

There exists at least one real $x$ for which $g'(x)=0$ where $x\in (a,b)$

I did not understand how I find minimum number of roots for $f(x)=0$ in $(a,b)$

Please have a look on my problem

edited Mar 19 '23 at 13:27

Thomas Andrews

177,126

asked Mar 19 '23 at 13:21

Priti Bisht

1

Reduce to the case $a=0, b=\pi$ just to make the notation easier. – Thomas Andrews Mar 19 '23 at 13:29
$f$ can be identically $0.$ – Anne Bauval Mar 19 '23 at 13:29
2

$g$ absolutely cannot be a function in $x$, since $x$ is a silent variable defined in the integrale – Zag Mar 19 '23 at 13:29
There is no question here. The sentence "Then number of ... is" might have been meant as a question, but if so, it should start with a question word like "What is the number of..." and end with a question mark (?) – Thomas Andrews Mar 19 '23 at 13:33
Moreover, this implicit question is different from the final one, which is about the minimum number. – Anne Bauval Mar 19 '23 at 13:37
4

Does this answer your question? Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $ – Anne Bauval Mar 19 '23 at 13:45
Also this: https://math.stackexchange.com/q/1286291/42969. – Martin R Mar 27 '23 at 07:27

Number of real solution of expression containing definite integral

0 Answers0