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The following questions focus on re-indexing sums.

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a)

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We 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 $n=\\simplify{k-{shift}}$.

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So we go through and change all the values or expressions of $n$ to $k$:

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• $n=\\var{start}\\implies k=\\simplify[!collectNumbers]{{start}+{shift}}=\\var{start+shift}$,
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• $n=\\var{stop}\\implies k=\\simplify[!collectNumbers]{{stop}+{shift}}=\\var{stop+shift}$,
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• The coefficient $t_n$ is now written as $t_{\\simplify{k-{shift}}}$ and $x^{\\simplify{n+{shift}}}$ is now written as $x^{k}$
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Therefore $\\displaystyle\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}=\\sum_{k=\\var{start+shift}}^\\var{stop+shift} t_{\\simplify{k-{shift}}}x^k$, but now we can choose to use $n$ instead of $k$ for our index (or dummy variable) so we replace $k$ with $n$:

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\$\\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\$

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b)

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We 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}\$

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Since 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\$

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but 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 $t_0$:

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\$y=t_0\\sum_{n=0}^\\infty \\frac{x^n}{n!\\var{c}^n}\$

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c)

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We 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}\$

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Since 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} \$

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but now we can choose to use $n$ instead of $k$ for our index (or dummy variable) so we replace $k$ with $n$:

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\$\\sinh(\\var{b}x)=\\sum_{n=0}^\\infty\\frac{\\var{b}^{2n+1}}{(2n+1)!}x^{2n+1}\$

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Often students need practice re-indexing series, these are three questions that give them that opportunity.

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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 $c$:

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\$\\sum_{n=\\var{start}}^{\\var{stop}}t_nx^{\\simplify{n+{shift}}}=\\sum_{n=a}^b t_cx^n\$

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where $a=$[], $b=$[] and $c=$[].

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Given \$y=\\sum_{n=0}^\\infty t_n x^n,\$

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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}\$

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rewrite the sum so that each value of the index corresponds to a non-zero coefficient:

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$\\displaystyle y=t_0\\sum_{n=0}^\\infty$ []

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Note: To input say $\\dfrac{10!5^{2n}}{x+1}$ you could type 10!*5^(2n)/(x+1)

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Given \$\\sinh(\\var{b}x)=\\sum_{n=0}^\\infty t_n x^n,\$

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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}\$

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rewrite the sum so that each value of the index corresponds to a non-zero coefficient:

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$\\displaystyle \\sinh(\\var{b}x)=\\sum_{n=0}^\\infty$ []

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Note: To input say $\\dfrac{10!5^{2n}}{x+1}$ you could type 10!*5^(2n)/(x+1)

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