Mixing
How was this function approximated?
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I found this passage in my book which is not very clear to me as to how it was obtained.
Given $ninmathbbN,,kin(0,1)$, then
$$(1+i)^n+kapprox(1-k)(1+i)^n+k(1+i)^n+1=(1+i)^n(1+ki)$$
calculus
asked 58 mins ago
user372003
218111
add a comment |Â
up vote
2
down vote
favorite
I found this passage in my book which is not very clear to me as to how it was obtained.
Given $ninmathbbN,,kin(0,1)$, then
$$(1+i)^n+kapprox(1-k)(1+i)^n+k(1+i)^n+1=(1+i)^n(1+ki)$$
calculus
asked 58 mins ago
user372003
218111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I found this passage in my book which is not very clear to me as to how it was obtained.
Given $ninmathbbN,,kin(0,1)$, then
$$(1+i)^n+kapprox(1-k)(1+i)^n+k(1+i)^n+1=(1+i)^n(1+ki)$$
calculus
asked 58 mins ago
user372003
218111
I found this passage in my book which is not very clear to me as to how it was obtained.
Given $ninmathbbN,,kin(0,1)$, then
$$(1+i)^n+kapprox(1-k)(1+i)^n+k(1+i)^n+1=(1+i)^n(1+ki)$$
calculus
calculus
asked 58 mins ago
user372003
218111
asked 58 mins ago
user372003
218111
asked 58 mins ago
user372003
218111
asked 58 mins ago
user372003
218111
asked 58 mins ago
user372003
218111
218111
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Is $i$ supposed to be small? If so, the derivative of $(1+i)^n$ with respect to $n$ is $(1+i)^nlog(1+i)$. Thus a first order approximation of $(1+i)^n+k$ is
$$
(1+i)^n+k(1+i)^nln(1+i).
$$
If $0<i<1$ then $iapproxln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^n+ik(1+i)^n$.
answered 46 mins ago
ThomasGrubb
9,93011335
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
add a comment |Â
up vote
3
down vote
Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)cdot 1 + xcdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
answered 42 mins ago
user7530
33.8k658110
add a comment |Â
up vote
1
down vote
Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^n+k$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
fracf(k)-f(0)k-0approxfracf(1)-f(0)1-0
$$
Rearranging this gives
$$
f(k)approx (1-k)f(0) + k f(1).
$$
answered 22 mins ago
grand_chat
18.5k11122
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Is $i$ supposed to be small? If so, the derivative of $(1+i)^n$ with respect to $n$ is $(1+i)^nlog(1+i)$. Thus a first order approximation of $(1+i)^n+k$ is
$$
(1+i)^n+k(1+i)^nln(1+i).
$$
If $0<i<1$ then $iapproxln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^n+ik(1+i)^n$.
answered 46 mins ago
ThomasGrubb
9,93011335
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
add a comment |Â
up vote
3
down vote
accepted
Is $i$ supposed to be small? If so, the derivative of $(1+i)^n$ with respect to $n$ is $(1+i)^nlog(1+i)$. Thus a first order approximation of $(1+i)^n+k$ is
$$
(1+i)^n+k(1+i)^nln(1+i).
$$
If $0<i<1$ then $iapproxln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^n+ik(1+i)^n$.
answered 46 mins ago
ThomasGrubb
9,93011335
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Is $i$ supposed to be small? If so, the derivative of $(1+i)^n$ with respect to $n$ is $(1+i)^nlog(1+i)$. Thus a first order approximation of $(1+i)^n+k$ is
$$
(1+i)^n+k(1+i)^nln(1+i).
$$
If $0<i<1$ then $iapproxln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^n+ik(1+i)^n$.
answered 46 mins ago
ThomasGrubb
9,93011335
Is $i$ supposed to be small? If so, the derivative of $(1+i)^n$ with respect to $n$ is $(1+i)^nlog(1+i)$. Thus a first order approximation of $(1+i)^n+k$ is
$$
(1+i)^n+k(1+i)^nln(1+i).
$$
If $0<i<1$ then $iapproxln(1+i)$ (again using a Taylor series approximation), and hence one arrives at the desired approximation of $(1+i)^n+ik(1+i)^n$.
answered 46 mins ago
ThomasGrubb
9,93011335
answered 46 mins ago
ThomasGrubb
9,93011335
answered 46 mins ago
ThomasGrubb
9,93011335
answered 46 mins ago
ThomasGrubb
9,93011335
9,93011335
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
add a comment |Â
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
1
1
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
@user372003 $n$ was a natural number, and I'm comfortable raising something to the power of a natural number. $k$ was assumed to be a small number, so we are essentially raising a number to the power of a natural number, with a small perturbation. This small change of $k$ tells me I should use a derivative to turn something nasty (raising something to the power $n+k$) into something I'm comfortable with (raising something to the power $n$)
– ThomasGrubb
34 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
but in order to expand shouldn’t you differentiate with respect to $k$ at $k=0$?
– user372003
32 mins ago
1
1
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
@user372003 We're basically talking about the same thing; I differentiated the function $(1+i)^n$ with respect to $n$, and taking $k$ to be a constant. You are talking about differentiating the function $(1+i)^n+k$, and taking $n$ to be a constant. You'll arrive at the same destination either way.
– ThomasGrubb
27 mins ago
add a comment |Â
up vote
3
down vote
Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)cdot 1 + xcdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
answered 42 mins ago
user7530
33.8k658110
add a comment |Â
up vote
3
down vote
Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)cdot 1 + xcdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
answered 42 mins ago
user7530
33.8k658110
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)cdot 1 + xcdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
answered 42 mins ago
user7530
33.8k658110
Some intuition to go with ThomasGrubb's thorough answer: for $a>1$, $a^x$ is an increasing function that interpolates between $1$ (at $x=0$) and $a$ (at $x=1$). A secant approximation of $a^x$ is ordinary linear interpolation of these extremes: $(1-x)cdot 1 + xcdot a.$
As ThomasGrubb notes, this approximation is increasingly poor as $a$ increases.
answered 42 mins ago
user7530
33.8k658110
answered 42 mins ago
user7530
33.8k658110
answered 42 mins ago
user7530
33.8k658110
answered 42 mins ago
user7530
33.8k658110
33.8k658110
add a comment |Â
add a comment |Â
up vote
1
down vote
Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^n+k$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
fracf(k)-f(0)k-0approxfracf(1)-f(0)1-0
$$
Rearranging this gives
$$
f(k)approx (1-k)f(0) + k f(1).
$$
answered 22 mins ago
grand_chat
18.5k11122
add a comment |Â
up vote
1
down vote
Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^n+k$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
fracf(k)-f(0)k-0approxfracf(1)-f(0)1-0
$$
Rearranging this gives
$$
f(k)approx (1-k)f(0) + k f(1).
$$
answered 22 mins ago
grand_chat
18.5k11122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^n+k$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
fracf(k)-f(0)k-0approxfracf(1)-f(0)1-0
$$
Rearranging this gives
$$
f(k)approx (1-k)f(0) + k f(1).
$$
answered 22 mins ago
grand_chat
18.5k11122
Notice that $k$ is assumed between $0$ and $1$. With $f(k):=(1+i)^n+k$, the approximation is that $f(k)$ grows roughly linearly as $k$ ranges from $0$ to $1$. Consequently,
$$
fracf(k)-f(0)k-0approxfracf(1)-f(0)1-0
$$
Rearranging this gives
$$
f(k)approx (1-k)f(0) + k f(1).
$$
answered 22 mins ago
grand_chat
18.5k11122
answered 22 mins ago
grand_chat
18.5k11122
answered 22 mins ago
grand_chat
18.5k11122
answered 22 mins ago
grand_chat
18.5k11122
18.5k11122
add a comment |Â
add a comment |Â
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