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

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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
by (160 points)
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.