To prove the function is strictly decreasing (nominal interest rate convertible $p$ times a year)
How would one prove that the function $f(p)=p[(1+i)^p-1]$ is strictly decreasing? Here $i>-1$.
I've tried the usual approach, i.e. finding the derivative of $f$ and got a pretty terrible expression $$f'(p)=(1+i)^{1/p}(1-frac 1p log(1+i))-1$$ and I cannot find where the derivative is positive and where it's negative?
Any help would be appreciated.
calculus actuarial-science
asked Nov 27 '18 at 20:30
windircurse
1,106820
add a comment |
How would one prove that the function $f(p)=p[(1+i)^p-1]$ is strictly decreasing? Here $i>-1$.
I've tried the usual approach, i.e. finding the derivative of $f$ and got a pretty terrible expression $$f'(p)=(1+i)^{1/p}(1-frac 1p log(1+i))-1$$ and I cannot find where the derivative is positive and where it's negative?
Any help would be appreciated.
calculus actuarial-science
asked Nov 27 '18 at 20:30
windircurse
1,106820
add a comment |
How would one prove that the function $f(p)=p[(1+i)^p-1]$ is strictly decreasing? Here $i>-1$.
I've tried the usual approach, i.e. finding the derivative of $f$ and got a pretty terrible expression $$f'(p)=(1+i)^{1/p}(1-frac 1p log(1+i))-1$$ and I cannot find where the derivative is positive and where it's negative?
Any help would be appreciated.
calculus actuarial-science
asked Nov 27 '18 at 20:30
windircurse
1,106820
How would one prove that the function $f(p)=p[(1+i)^p-1]$ is strictly decreasing? Here $i>-1$.
I've tried the usual approach, i.e. finding the derivative of $f$ and got a pretty terrible expression $$f'(p)=(1+i)^{1/p}(1-frac 1p log(1+i))-1$$ and I cannot find where the derivative is positive and where it's negative?
Any help would be appreciated.
calculus actuarial-science
calculus actuarial-science
asked Nov 27 '18 at 20:30
windircurse
1,106820
asked Nov 27 '18 at 20:30
windircurse
1,106820
edited Nov 27 '18 at 20:43
asked Nov 27 '18 at 20:30
windircurse
1,106820
asked Nov 27 '18 at 20:30
windircurse
1,106820
asked Nov 27 '18 at 20:30
windircurse
1,106820
1,106820
add a comment |
add a comment |
1 Answer
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If $h$ is increasing and $g$ is increasing, then $hcdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)cdot g(x)>h(y)cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $hcdot g$ is decreasing by looking at it as $-(hcdot -g)$ since $-g$ would be positive and increasing.
Assume now for the moment that we are working on the domains $(0,infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.
If $i>0$, $g$ is increasing and positive. Thus $hcdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $hcdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $hcdot g$ will be zero.
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
add a comment |
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1 Answer
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If $h$ is increasing and $g$ is increasing, then $hcdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)cdot g(x)>h(y)cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $hcdot g$ is decreasing by looking at it as $-(hcdot -g)$ since $-g$ would be positive and increasing.
Assume now for the moment that we are working on the domains $(0,infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.
If $i>0$, $g$ is increasing and positive. Thus $hcdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $hcdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $hcdot g$ will be zero.
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
add a comment |
If $h$ is increasing and $g$ is increasing, then $hcdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)cdot g(x)>h(y)cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $hcdot g$ is decreasing by looking at it as $-(hcdot -g)$ since $-g$ would be positive and increasing.
Assume now for the moment that we are working on the domains $(0,infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.
If $i>0$, $g$ is increasing and positive. Thus $hcdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $hcdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $hcdot g$ will be zero.
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
add a comment |
If $h$ is increasing and $g$ is increasing, then $hcdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)cdot g(x)>h(y)cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $hcdot g$ is decreasing by looking at it as $-(hcdot -g)$ since $-g$ would be positive and increasing.
Assume now for the moment that we are working on the domains $(0,infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.
If $i>0$, $g$ is increasing and positive. Thus $hcdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $hcdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $hcdot g$ will be zero.
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
If $h$ is increasing and $g$ is increasing, then $hcdot g$ is increasing if for all $x$, $h(x),g(x)> 0$. Indeed, if $x> y$, we have $h(x)cdot g(x)>h(y)cdot g(y)$ by the increasing properties.
Similarly, if $h$ is increasing and positive and $g$ is decreasing and negative, then $hcdot g$ is decreasing by looking at it as $-(hcdot -g)$ since $-g$ would be positive and increasing.
Assume now for the moment that we are working on the domains $(0,infty)$ and take $h(p) = p$ and $g(p) = (1+i)^p - 1$.
If $i>0$, $g$ is increasing and positive. Thus $hcdot g$ will be increasing. If $-1>i>0$, $g$ is decreasing and negative and thus $hcdot g$ will be decreasing. If $i=0$, $g$ is zero and thus $hcdot g$ will be zero.
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
edited Nov 27 '18 at 21:08
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
answered Nov 27 '18 at 20:59
Stan Tendijck
1,411210
1,411210
add a comment |
add a comment |
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