Rewrite the sum \\[\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}\\] so that it only involves $x^{n}$ terms, by specifying values for $a$ and $b$ and an expression for $c$:
\n\\[\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}=\\sum_{n=a}^b t_cx^n\\]
\nwhere $a=$[[0]], $b=$[[1]] and $c=$[[2]].
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\nwhere \\[t_n=\\begin{cases}0,& \\text{if } n \\text{ is not divisible by } \\var{d}\\\\\\dfrac{t_0}{\\left(\\frac{n}{\\var{d}}\\right)! \\var{c}^\\frac{n}{\\var{d}}},& \\text{if } n \\text{ is divisible by }\\var{d}\\end{cases}\\]
\nrewrite the sum so that each value of the index corresponds to a non-zero coefficient:
\n\n$\\displaystyle y=t_0\\sum_{n=0}^\\infty$ [[0]]
\nNote: To input say $\\dfrac{10!5^{2n}}{x+1}$ you could type 10!*5^(2n)/(x+1)
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\nwhere \\[t_n=\\begin{cases}\\frac{\\var{b}^n}{n!},& \\text{if } n \\text{ is odd}\\\\0,& \\text{if } n \\text{ is even}\\end{cases}\\]
\nrewrite the sum so that each value of the index corresponds to a non-zero coefficient:
\n\n$\\displaystyle \\sinh(\\var{b}x)=\\sum_{n=0}^\\infty$ [[0]]
\nNote: To input say $\\dfrac{10!5^{2n}}{x+1}$ you could type 10!*5^(2n)/(x+1)
"}], "functions": {}, "name": "Series: re-indexing", "ungrouped_variables": ["start", "length", "stop", "shift", "b", "c", "d"], "preamble": {"js": "", "css": ""}, "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "Often students need practice re-indexing series, these are three questions that give them that opportunity.
"}, "extensions": [], "statement": "The following questions focus on re-indexing sums.
", "advice": "a)
\nWe have $\\displaystyle\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}$ but we want $x^{n}$ terms! It's like we want to replace $\\simplify{n+{shift}}$ with $n$ but it can get a bit messy or confusing if we do this because we have to keep track of which are the 'old' and which are the 'new' $n$'s. Instead, we normally do the substitution $k=\\simplify{n+{shift}}$ which then implies $n=\\simplify{k-{shift}}$.
\nSo we go through and change all the values or expressions of $n$ to $k$:
\nTherefore $\\displaystyle\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}=\\sum_{k=\\var{start+shift}}^\\var{stop+shift} t_{\\simplify{k-{shift}}}x^k$, but now we can choose to use $n$ instead of $k$ for our index (or dummy variable) so we replace $k$ with $n$:
\n\\[\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}=\\sum_{n=\\var{start+shift}}^\\var{stop+shift} t_{\\simplify{n-{shift}}}x^n\\]
\n\nb)
\nWe have $y=\\sum_{n=0}^\\infty t_n x^n$ where \\[t_n=\\begin{cases}0,& \\text{if } n \\text{ is not divisible by } \\var{d}\\\\\\dfrac{t_0}{\\left(\\frac{n}{\\var{d}}\\right)! \\var{c}^\\frac{n}{\\var{d}}},& \\text{if } n \\text{ is divisible by }\\var{d}\\end{cases}\\]
\nSince a lot of the terms in this sum are actually just zero and the sum relies on a piecewise function, it would be nice to make this sum only deal with the non-zero terms. Notice the only non-zero terms are when $n=0, \\,\\var{d}, \\,\\var{2*d}, \\,\\var{3*d}, \\ldots$ (the multiples of $\\var{d}$) so we make the substitution $k=\\frac{n}{\\var{d}}$, which will make the non-zero terms correspond to $k=0,\\,1,\\,2,\\,3,\\ldots$. So now our sum will be \\[y=\\sum_{k=0}^\\infty \\frac{t_0}{k!\\var{c}^k} x^k\\]
\nbut now we can choose to use $n$ instead of $k$ for our index (or dummy variable) so we replace $k$ with $n$ and we can pull out the common factor of $t_0$:
\n\\[y=t_0\\sum_{n=0}^\\infty \\frac{x^n}{n!\\var{c}^n}\\]
\n\nc)
\nWe have $\\sinh(\\var{b}x)=\\sum_{n=0}^\\infty t_n x^n$ where \\[t_n=\\begin{cases}\\frac{\\var{b}^n}{n!},& \\text{if } n \\text{ is odd}\\\\0,& \\text{if } n \\text{ is even}\\end{cases}\\]
\nSince half the terms in this sum are actually just zero and the sum relies on a piecewise function, it would be nice to make this sum only deal with the non-zero terms. Notice the only non-zero terms are when $n=1,\\,3,\\,5, \\ldots$ so we make the substitution $n=2k+1$, which will make the non-zero terms correspond to $k=0,\\,1,\\,2,\\,3,\\ldots$. So now our sum will be \\[\\sum_{k=0}^\\infty\\frac{\\var{b}^{2k+1}}{(2k+1)!}x^{2k+1} \\]
\nbut now we can choose to use $n$ instead of $k$ for our index (or dummy variable) so we replace $k$ with $n$:
\n\\[\\sinh(\\var{b}x)=\\sum_{n=0}^\\infty\\frac{\\var{b}^{2n+1}}{(2n+1)!}x^{2n+1}\\]
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