What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Power Series and MatricesInterval of Converge for a Power SeriesWhen is a Power Series a Geometric Series?Power Series Comprehension$(x-x_0)^0$ in power seriesOn the composition of formal power seriesComposition of Convergent Power Series is a Convergent Power SeriesUnderstanding ProofWiki's proof of differentiation of sine power seriesSubtracting two power seriesPower series: show that c_n=0 for all natural numbers
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What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Power Series and MatricesInterval of Converge for a Power SeriesWhen is a Power Series a Geometric Series?Power Series Comprehension$(x-x_0)^0$ in power seriesOn the composition of formal power seriesComposition of Convergent Power Series is a Convergent Power SeriesUnderstanding ProofWiki's proof of differentiation of sine power seriesSubtracting two power seriesPower series: show that c_n=0 for all natural numbers
$begingroup$
What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$
Examples of power series that I've seen in my textbook:
$sum_n=0^infty x^n$
$sum_n=0^infty n!x^n$
$sum_n=0^infty (-1)^nx^2n/2^2n(n!)^2$
$sum_n=0^infty (-3)^nx^n/sqrtn+1$
$sum_n=0^infty n(x+2)^n/3^n+1$
The only one that conforms to the equation of $c_n(x-a)^n$ is the last one, in which I suppose $a=-2$ and $c_n=n/3^n+1$
Are those the correct values of $a$ and $c_n$ for the last series I gave?
Also, why do all the other series seem to not take on the form of $c_n(x-a)^n$ in that there is no $-a$ term within brackets along with the $x$ (it's usually just the $x$)?
I just don't think I'm understanding what a power series is on a conceptual level and I think my confusion stems from this equation. If anyone could help me better understand this I'd appreciate it a lot!
calculus sequences-and-series power-series
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
$endgroup$
add a comment |
$begingroup$
What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$
Examples of power series that I've seen in my textbook:
$sum_n=0^infty x^n$
$sum_n=0^infty n!x^n$
$sum_n=0^infty (-1)^nx^2n/2^2n(n!)^2$
$sum_n=0^infty (-3)^nx^n/sqrtn+1$
$sum_n=0^infty n(x+2)^n/3^n+1$
The only one that conforms to the equation of $c_n(x-a)^n$ is the last one, in which I suppose $a=-2$ and $c_n=n/3^n+1$
Are those the correct values of $a$ and $c_n$ for the last series I gave?
Also, why do all the other series seem to not take on the form of $c_n(x-a)^n$ in that there is no $-a$ term within brackets along with the $x$ (it's usually just the $x$)?
I just don't think I'm understanding what a power series is on a conceptual level and I think my confusion stems from this equation. If anyone could help me better understand this I'd appreciate it a lot!
calculus sequences-and-series power-series
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
$endgroup$
5
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06
add a comment |
$begingroup$
What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$
Examples of power series that I've seen in my textbook:
$sum_n=0^infty x^n$
$sum_n=0^infty n!x^n$
$sum_n=0^infty (-1)^nx^2n/2^2n(n!)^2$
$sum_n=0^infty (-3)^nx^n/sqrtn+1$
$sum_n=0^infty n(x+2)^n/3^n+1$
The only one that conforms to the equation of $c_n(x-a)^n$ is the last one, in which I suppose $a=-2$ and $c_n=n/3^n+1$
Are those the correct values of $a$ and $c_n$ for the last series I gave?
Also, why do all the other series seem to not take on the form of $c_n(x-a)^n$ in that there is no $-a$ term within brackets along with the $x$ (it's usually just the $x$)?
I just don't think I'm understanding what a power series is on a conceptual level and I think my confusion stems from this equation. If anyone could help me better understand this I'd appreciate it a lot!
calculus sequences-and-series power-series
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
$endgroup$
What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$
Examples of power series that I've seen in my textbook:
$sum_n=0^infty x^n$
$sum_n=0^infty n!x^n$
$sum_n=0^infty (-1)^nx^2n/2^2n(n!)^2$
$sum_n=0^infty (-3)^nx^n/sqrtn+1$
$sum_n=0^infty n(x+2)^n/3^n+1$
The only one that conforms to the equation of $c_n(x-a)^n$ is the last one, in which I suppose $a=-2$ and $c_n=n/3^n+1$
Are those the correct values of $a$ and $c_n$ for the last series I gave?
Also, why do all the other series seem to not take on the form of $c_n(x-a)^n$ in that there is no $-a$ term within brackets along with the $x$ (it's usually just the $x$)?
I just don't think I'm understanding what a power series is on a conceptual level and I think my confusion stems from this equation. If anyone could help me better understand this I'd appreciate it a lot!
calculus sequences-and-series power-series
calculus sequences-and-series power-series
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
asked Apr 1 at 19:09
James RonaldJames Ronald
3648
3648
5
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06
add a comment |
5
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06
5
5
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.
Given a power series $sum_n=0^infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)
So if your last example $sum_n=0^infty fracn(x+2)^n3^n+1$ should look like the power seris $sum_n=0^infty fracnx^n3^n+1$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.
answered Apr 1 at 19:32
kccukccu
11.4k11231
$endgroup$
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.
Given a power series $sum_n=0^infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)
So if your last example $sum_n=0^infty fracn(x+2)^n3^n+1$ should look like the power seris $sum_n=0^infty fracnx^n3^n+1$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.
answered Apr 1 at 19:32
kccukccu
11.4k11231
$endgroup$
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
add a comment |
$begingroup$
Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.
Given a power series $sum_n=0^infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)
So if your last example $sum_n=0^infty fracn(x+2)^n3^n+1$ should look like the power seris $sum_n=0^infty fracnx^n3^n+1$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.
answered Apr 1 at 19:32
kccukccu
11.4k11231
$endgroup$
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
add a comment |
$begingroup$
Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.
Given a power series $sum_n=0^infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)
So if your last example $sum_n=0^infty fracn(x+2)^n3^n+1$ should look like the power seris $sum_n=0^infty fracnx^n3^n+1$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.
answered Apr 1 at 19:32
kccukccu
11.4k11231
$endgroup$
Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.
Given a power series $sum_n=0^infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)
So if your last example $sum_n=0^infty fracn(x+2)^n3^n+1$ should look like the power seris $sum_n=0^infty fracnx^n3^n+1$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.
answered Apr 1 at 19:32
kccukccu
11.4k11231
answered Apr 1 at 19:32
kccukccu
11.4k11231
answered Apr 1 at 19:32
kccukccu
11.4k11231
answered Apr 1 at 19:32
kccukccu
11.4k11231
11.4k11231
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
add a comment |
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
$begingroup$
My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:05
add a comment |
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5
$begingroup$
The others also conform to that form, only $a=0$.
$endgroup$
– Arthur
Apr 1 at 19:10
$begingroup$
Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you!
$endgroup$
– James Ronald
Apr 1 at 20:06