Question

$If \: |a_1sinx +a_2sin2x + a_3sin3x + - - - + a_nsinnx | \leqslant |sinx| \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ for \: all \: x \in \: real \: numbers \: then \: prove \: that \: |a_1 + 2a_2 + 3a_3 + - - + na_n| \leqslant 1$
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Answer 1

Answer:

\: \: \: \: \: \\ for \: all \: x \in \: real \: numbers \: then \: prove \: that \: |a_1 + 2a_2 + 3a_3 + - - + na_n| \leqslant 1[/tex]

Please don't spam.

Quality answer needed.

Moderators, Brainly stars , genius, please answer as earliest.

Thanks for all your support and cooperation.

Helps me alot in my studies.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: |a_1sinx +a_2sin2x + a_3sin3x + - - - + a_nsinnx | \leqslant |sinx|$

Let we assume that,

$\sf \: f(x) = a_1sinx +a_2sin2x + a_3sin3x + - - - + a_nsinnx$

$\purple{\bf \implies\: |f(x)| \leqslant |sinx| }$

On substituting x = 0, we get

$\rm :\implies\: |f(0)| \leqslant |sin0|$

$\rm :\implies\: |f(0)| \leqslant |0|$

$\rm :\implies\: |f(0)| \leqslant 0$

$\rm :\implies\: |f(0)| = 0$

$\bf\implies \:f(0) = 0 - - - (1)$

Also, as assume above

$\sf \: f(x) = a_1sinx +a_2sin2x + a_3sin3x + - - - + a_nsinnx$

On differentiating both sides w. r. t x, we get

$\sf \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}\bigg [a_1sinx +a_2sin2x + a_3sin3x + - - - + a_nsinnx\bigg]$

$\sf \: f'(x) = a_1cosx +2a_2cos2x + 3a_3cos3x + - - - + na_ncosnx$

On substituting x = 0, we get

$\red{\rm :\longmapsto\: \: f'(0) = a_1 +2a_2 + 3a_3 + - - - + na_n }$

So,

$\red{\rm :\longmapsto\: \: |f'(0)| = \bigg | a_1 +2a_2 + 3a_3 + - - - + na_n \bigg |} - - - (1)$

Now, We know that

By definition of differentiation,

$\rm :\longmapsto\:f'(0) = \displaystyle\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

$\rm :\longmapsto\:f'(0) = \displaystyle\lim_{x \to 0} \frac{f(x) - 0}{x}$

$\red{\bigg \{ \because \:f(0) = 0 \bigg \}}$

$\rm :\longmapsto\:f'(0) = \displaystyle\lim_{x \to 0} \frac{f(x)}{x}$

So,

$\rm :\longmapsto\: | f'(0) |= \displaystyle\lim_{x \to 0}\bigg | \frac{f(x)}{x}\bigg |$

$\rm :\longmapsto\: | f'(0) |= \displaystyle\lim_{x \to 0}\bigg | \frac{f(x)}{x}\bigg | \leqslant \displaystyle\lim_{x \to 0}\bigg | \frac{sinx}{x}\bigg |$

$\purple{ \{\bf \because\: |f(x)| \leqslant |sinx| \} }$

$\rm :\longmapsto\: | f'(0) |\leqslant \displaystyle\lim_{x \to 0}\bigg | \frac{sinx}{x}\bigg |$

We know,

$\boxed{ \bf{ \: \displaystyle\lim_{h \to 0} \frac{sinh}{h} = 1}}$

So, using this,

$\rm :\longmapsto\: | f'(0) |\leqslant 1$

Hence,

From equation (1), we get

$\red{\rm :\longmapsto\: \: \bigg | a_1 +2a_2 + 3a_3 + - - - + na_n \bigg | \leqslant 1}$

Hence, Proved