how do you prove that the nominal interest rate i^p is a decreasing function of p?

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We can prove that \(i^{(p)}\) is a decreasing function of \(p\) as follows:

Let \(f(x)=x(e^{\delta/x}-1\))

Then \(f'(x)=e^{\delta/x}(1-{\delta/x})-1\)

We know from elementary calculus that:

\(e^y(1-y)-1<0\) for all \(y\neq0\)

Thus \(f'(x)<0\) \( \forall \space x>0 \)

\(f(x)\) is thus, a decreasing function on \((0,\infty)\) ad for each m, 1, 2, 3

\(f(m+1)<f(m)\)

Hence: \(i^{(m+1)}<i^{(m)}\) for each m=1,2,3..

Similarly we can prove that \(d<d^{(2)}<d^{(3)}\)

Let \(g(x)=x(1-e^{-\delta/x})\)

\(g'(x)=1-e^{-\delta/x}(1 + {\delta/x})\)

\(=-[e^y(1-y)-1]\)

where \(y=\delta/x \)

Then we have that \(g'(x)>0\) \(\forall x>0\) and hence that \(g(x)\) is an increasing function \((0,\infty)\)

Hence \(d^{(m+1)}>d^{(m)}\) for each m=1,2..

+1 vote

Rearranging (1+i) = \( {(1+\frac{{i}^{(p)}}{p})}^{p} \), we get that:

$${i}^{(p)} = p[{(1+i)^{\frac{1}{p}}} -1] $$

\( \frac{1}{p} \) is a decreasing function of p.

(1+i) > 1 assuming i > 0.

Thus, \( {(1+i)}^{\frac{1}{p}} \) is a decreasing function of p and so [\( {(1+i)}^{\frac{1}{p}} \) - 1] is also a decreasing function of p.

Now, p increases slower than [\( {(1+i)}^{\frac{1}{p}} \) - 1] decreases; therefore, \( {i}^{(p)} \) = p[\( {(1+i)}^{\frac{1}{p}} \) - 1] is a decreasing function of p.

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