Induction Proof of trig inequality $\sum_{k=0}^n |\cos k| \ge \frac n2$

Question

This is for a course, I don't want the answer just a prod in the right direction!

I've got a problem that states

let n be an integer such that $$n\gt0$$ $$\text{Prove: }\sum_{k=0}^n |\cos k| \ge \frac n2$$

I'm using induction to prove this. First I showed my base case:

When $P(0)$ $\sum_{k=0}^0 |\cos o| \ge \frac 02$

Which simplifies down into $|1|\ge 0$ which holds true. So I moved to the induction step.

Assume: $P(n)$ is true such that $\sum_{k=0}^n |\cos k| \ge \frac n2$

WTS: P(n+1) is true such that $\sum_{k=0}^{n+1} |\cos k| \ge \frac n2$

So I broke the summation up into two parts

$\sum_{k=0}^{n+1} |\cos k|=\sum_{k=0}^n |\cos k|+ \cos(n+1)$ Which I can substitute back in and get

$\sum_{k=0}^n |\cos k|+ \cos(n+1) \ge \frac {n+1}2$

I know, well I think, that this is where I use my Induction Hypothesis $\sum_{k=0}^n |\cos k| \ge \frac n2$

But I don't now how to show my next step.

Can I just substitue $\frac n2$ in for $\sum_{k=0}^n |\cos k|$ and continue to solve? or do I have to show more steps in the middle? or am I completely lost? If anyone can prod me into the right direction it would be much appreciated! Thanks for your help!

First Edit @Bungo

This is the way I've been doing it. Is my logic wrong? Do I need to prove the lower bound? Or is my logic okay?

So I broke the sum down and re wrote it as $\sum_{k=0}^n + |\cos n+1| \ge \frac {n+1}2$

And then substituted from the induction hypothesis and wrote it as $\frac n2 + |\cos n+1| \ge \frac {n+1}2$

Then I simplified and got $|\cos n+1| \ge \frac 12$

So now I'm trying to prove that statement as true. Is this okay? Or should I go back and try with the triplets as you suggested? Thanks!

Answer 1

The statement $$\sum_{k=0}^{n}|\cos(k)| \geq \frac{n}{2}$$ appears to be true, but it is not in general true that $$\frac{n}{2} + |\cos(n+1)| \geq \frac{n+1}{2}$$ (e.g., this is false for $n=1$). So I don't think it will be useful to decompose the sum as $$\sum_{k=0}^{n+1} |\cos(k)| = |\cos(n+1)| + \sum_{k=0}^n |\cos(k)|$$ because we don't have a lower bound for $|\cos(n+1)|$ other than zero.

Instead, one might look at pairs of the form $|\cos(n)| + |\cos(n+1)|$ and try to find a useful lower bound for these. Plotting $|\cos(x)| + |\cos(x+1)|$ at Wolfram Alpha suggests that $|\cos(x)| + |\cos(x+1)| > 0.8$ for all $x$, which is still not good enough: we need a lower bound of $1$ if we add in pairs.

So let's look at triplets: $|\cos(n-1)| + |\cos(n)| + |\cos(n+1)|$. Wolfram Alpha's plot shows that the function $|\cos(x-1)| + |\cos(x)| + |\cos(x+1)|$ stays above $1.5$, so if you can prove that lower bound, breaking the sum into groups of triplets should suffice. I think proving the lower bound will be grungy, though. Hopefully someone can propose a nicer solution.

Answer 2

Usually in Induction we do the following:

• Show that the relation is true for the case $k=1$, i.e., prove that $\mathfrak{R}(1)$ is true, usually by putting values.
• Assume truth of the relation for a general value of $k$, i.e., assume $\mathfrak{R}(k)$ is true.
• Show that by assuming truth for a value of $k$ proves that the relation is also true for $(k+1)$, usually by showing that $\mathfrak{R}(k+1)=f(\mathfrak{R}(k))$* and then putting the expression of $\mathfrak{R}(k)$ into this and then establishing truth of $\mathfrak{R}(k+1)$.
• If this is done, you have proved the theorem/identity by induction.

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[*or in simple words, express $\mathfrak{R}(k+1)$ in terms of $\mathfrak{R}(k)$, let this correlation be denoted by $f$, usually but not always you get $\mathfrak{R}(k+1)=\mathfrak{R}(k)+\lambda$ or $f=\mathfrak{R}(k)+\lambda$]

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Trying to use induction in this case:

• $P(1)$ is true.
• Assuming $P(n)$ is true such that $\sum_{k=0}^n |\cos k| \ge \frac n2$
• Now trying to prove that $P(n+1)$ is true using $P(n)$, or: $$P(n+1)=P(n)+|\cos(n+1)|\ge\frac n2+|\cos(n+1)|\ge^*\frac{n+1}2$$ The $^*$ relation is not always true, so it's not possible to prove by this simple induction, you have to evoke something more complex.

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If you are not bound to use induction you may use the result: $$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos\left(\alpha+(n-1)\beta\right)\\=\sum_{k=0}^{n-1}\cos\left(\alpha+k\beta\right)=\frac{\cos\left(\alpha+\frac{n-1}2\beta\right)\sin\left(\frac{n\beta}2\right)}{\sin\left(\frac{\beta}2\right)}$$