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If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way to simplify $f(x)$. What is it?

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  • $\begingroup$ may be you mean in numerator $\cos 5x+5\cos 3x+10 \cos x$ $\endgroup$
    – juantheron
    Jun 21 '16 at 8:01
  • $\begingroup$ $f(0)=\frac7{32}$ and as the function is even, $f'(0)=0$. The rest is up to you :) $\endgroup$
    – user65203
    Jun 21 '16 at 8:33
  • $\begingroup$ The final answer is coming to be (-11/16), isn't it? Please, somebody confirm. $\endgroup$ Jun 22 '16 at 10:42
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we can simplify the fraction as $$\frac{2\cos3x\cos2x+5\cos3x}{2\cos^23x-1+6[2\cos3x\cos x]+9\cos2x+10}$$ $$=\frac{(2\cos2x+5)\cos3x}{2\cos^23x+12\cos3x\cos x+18\cos^2x}$$ $$=\frac{(2\cos2x+5)\cos3x}{2(\cos3x+3\cos x)^2}$$ $$=\frac{[2(2c^2-1)+5](4c^3-3c)}{2(4c^3)^2}$$ $$=\frac{(4c^2+3)(4c^2-3)}{32c^5}$$ $$=\frac 12\sec x-\frac{9}{32}\sec^5x$$

Now you can differentiate this twice quite easily

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  • $\begingroup$ This is a nice simplification, indeed. $\endgroup$ Jun 21 '16 at 9:59
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$f(x) = \frac{u(x)}{v(x)}$

so $f(x) v(x) = u(x)$

Differentiating gives: $f'(x) v(x) + v'(x) f(x) = u'(x)$

Differentiating again: $f''(x) v(x) + f'(x) v'(x) + v''(x)f(x)+f'(x)v'(x) = u''(x)$

Setting $x=0$ and noticing $u'(0)=v'(0)=0$ (only $\sin$ terms appear in the derivatives) gives:

$f'(0)v(0)=0 \Rightarrow f'(0)=0$

Then $f''(0)v(0)+v''(0)f(0)=u''(0)$ and finally:

$f''(0)=\frac{u''(0)-v''(0)f(0)}{v(0)}$

I'll let you compute $f(0)$,$v(0)$, $u''(0)$ and $v''(0)$ which is elementary.

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A "simple" way could be to consider Taylor series; start using $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ and replace $y$ by what is required.

Doing it, the numerator should be $$7-\frac{71 x^2}{2}+\frac{1031 x^4}{24}+O\left(x^6\right)$$ and the denominator $$32-96 x^2+128 x^4+O\left(x^6\right)$$ Now, long division.

When you get the result, remember that $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ Just identify the terms to get $f(0)$, $f'(0)$ and $f''(0)$.

Edit

As Yves Daoust commented, it will be sufficient to use $$\cos(y)=1-\frac{y^2}{2}+O\left(y^4\right)$$ This makes the work much faster.

Update

We may suppose that, for this problem, the coefficients were selected in order to allow nice trigonometric simplifications such as David Quinn's ones.

Let us take the most general case of $$f(x)=\frac{\sum_{i=0}^m a_i \cos(i x)}{\sum_{i=0}^n b_i \cos(i x)}$$ Using the same approach as in this answer we shall have $$f(x)=\frac{A_0+A_1 x^2+O\left(x^4\right)}{B_0+B_1 x^2+O\left(x^4\right)}=\frac{A_0}{B_0}+\frac{A_1B_0-A_0B_1}{B_0^2}x^2+O\left(x^4\right)$$ using $$A_0=\sum _{i=0}^m a_i\qquad A_1=-\frac{1}{2} \sum _{i=0}^m i^2 a_i\qquad B_0=\sum _{i=0}^n b_i\qquad B_1=-\frac{1}{2} \sum _{i=0}^n i^2 b_i\qquad$$

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  • $\begingroup$ Hi Claude. Excellent approach! Isn't it enough to develop to the quadratic term ? And as we will get $(a+bx^2)/(c+dx^2)$, this yields $a/c(1+(b/a-d/c)x^2)$. $\endgroup$
    – user65203
    Jun 21 '16 at 8:39
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    $\begingroup$ @YvesDaoust. Thank you. I suppose it is. I have the bad tendency to always use one extra term just in case of possible cancellations. $\endgroup$ Jun 21 '16 at 8:44
  • $\begingroup$ I know what you mean :) $\endgroup$
    – user65203
    Jun 21 '16 at 8:57
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There doesn't appear to be a way to simplify f(x) to make it easy to differentiate, however one way to make the differentiation less labour intensive is to express $f'(x)$ and $f''(x)$ in terms of the derivatives of the numerator and denominator.

Beginning from $f(x)=\frac{u(x)}{v(x)}$

Using the quotient rule

$f'(x)=\frac{vu'-uv'}{v^2}$

Applying the quotient, product, and chain rules

$f''(x)=\frac{v^2(v'u'+vu''-v'u'-uv'')-2vv'(vu'-uv')}{v^4}$

If you calculate $u(0), u'(0),u''(0)$ and $v(0), v'(0),v''(0)$ then it is straightforward to find the second derivative of $f$

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Try to simplify the denominator of f(x):

Note that by the identity $\cos(2t)=2\cos^2t-1$ $$ \cos(3t)+6\cos(2t)+15\cos t+10=4\cos^3t+12\cos^2t+12\cos t+4=4(\cos t+1)^3 $$ so $$ \cos(6x)+6\cos(4x)+15\cos(2x)+10=4(\cos(2x)+1)^3 $$ Therefore $$ f(x)=\frac{1}{4}\cdot\overbrace{\frac{\cos x}{(\cos(2x)+1)^3}}^{(1)}+\frac{5}{4}\cdot\overbrace{\frac{\cos(3x)}{(\cos(2x)+1)^3}}^{(2)}+\frac{1}{4}\cdot\overbrace{\frac{\cos(5x)}{(\cos(2x)+1)^3}}^{(3)} $$ Now, in order to compute $f'(x)$ and $f''(x)$ compute $(i)'$ and $(i)''$ for each $i\in\{1,2,3\}$. Then assign $x=0$.

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We can try to simplify the fraction using trigonometric formulae :

$$\cos (x) + 5\cos (3x) + \cos (5x)=\cos(x) (2 \cos(2 x)-1) (2 \cos(2 x)+5)$$ and $$\cos (6x) + 6\cos(4x) + 15\cos(2x) + 10=32 \cos^6(x)$$

So $$f(x)=\frac{(2 \cos(2 x)-1) (2 \cos(2 x)+5)}{32\cos^5(x)}=\frac{4\cos^2(2x)+8\cos(2x)-5}{32\cos^5(x)}$$ $$f'(x)=\frac{1}{32}(5 (12 \cos^2(2 x)-5) \tan(x) \sec^5(x)-48 \sin(2 x) \cos(2 x) \sec^5(x))\\=\frac{1}{32}\sec^5(x)(5 (12 \cos^2(2 x)-5) \tan(x)-48 \sin(2 x) \cos(2 x) )$$

But the calculation of $f''(x)$ is still pretty tedious...

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