Evaluate...
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I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$
I do as follow
$sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$
I couldn't able to get the right hand
Any help will be appreciated! Thanks
calculus summation
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$begingroup$
I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$
I do as follow
$sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$
I couldn't able to get the right hand
Any help will be appreciated! Thanks
calculus summation
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add a comment |
$begingroup$
I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$
I do as follow
$sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$
I couldn't able to get the right hand
Any help will be appreciated! Thanks
calculus summation
$endgroup$
I'd like to Prove that $sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=left(sumlimits_{n=0}^{infty}frac{1}{n!}a^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}b^nright)$
I do as follow
$sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{r=0}^{0}left(frac{1}{(0-r)!}a^{0-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{1}left(frac{1}{(1-r)!}a^{1-r}right)left(frac{1}{r!}b^{r}right)+sumlimits_{r=0}^{2}left(frac{1}{(2-r)!}a^{2-r}right)left(frac{1}{r!}b^{r}right)+cdots$
I couldn't able to get the right hand
Any help will be appreciated! Thanks
calculus summation
calculus summation
asked Dec 17 '18 at 6:53
user62498user62498
1,948613
1,948613
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3 Answers
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$begingroup$
You might find it easier to start from the RHS and show that
$$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$
Actually, for me this is the only step. It's just how you multiply two series.
Something more interesting would be to see what you can make of
$$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$
and the Binomial Theorem.
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$begingroup$
Hint
Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.
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I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
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– user62498
Dec 17 '18 at 7:37
add a comment |
$begingroup$
By hint Claude Leibovici,
$ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand
$e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$
Please let me know if this makes sense to you
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Your Answer
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
You might find it easier to start from the RHS and show that
$$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$
Actually, for me this is the only step. It's just how you multiply two series.
Something more interesting would be to see what you can make of
$$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$
and the Binomial Theorem.
$endgroup$
add a comment |
$begingroup$
You might find it easier to start from the RHS and show that
$$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$
Actually, for me this is the only step. It's just how you multiply two series.
Something more interesting would be to see what you can make of
$$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$
and the Binomial Theorem.
$endgroup$
add a comment |
$begingroup$
You might find it easier to start from the RHS and show that
$$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$
Actually, for me this is the only step. It's just how you multiply two series.
Something more interesting would be to see what you can make of
$$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$
and the Binomial Theorem.
$endgroup$
You might find it easier to start from the RHS and show that
$$ left(sum_{n=0}^{infty}frac{1}{n!}a^nright)left(sum_{n=0}^{infty}frac{1}{n!}b^nright) = sumlimits_{n=0}^{infty}sum_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right). $$
Actually, for me this is the only step. It's just how you multiply two series.
Something more interesting would be to see what you can make of
$$ frac{1}{(n-r)!r!}a^{n - r}b^r = frac{1}{n!} binom{n}{r} a^{n - r}b^r $$
and the Binomial Theorem.
answered Dec 17 '18 at 7:12
Trevor GunnTrevor Gunn
14.6k32046
14.6k32046
add a comment |
add a comment |
$begingroup$
Hint
Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.
$endgroup$
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
add a comment |
$begingroup$
Hint
Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.
$endgroup$
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
add a comment |
$begingroup$
Hint
Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.
$endgroup$
Hint
Consider the inner summation and show that $$sumlimits_{r=0}^{n}left(frac{a^{n-r}}{(n-r)!}right)left(frac{b^{r}}{r!}right)=frac{(a+b)^n}{n!}$$ Now, the lhs will be very familiar to you and the remaining is simple.
answered Dec 17 '18 at 7:15
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
add a comment |
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
$begingroup$
I do it how can I get $sum_{n=0}^{infty}frac{(a+b)^n}{n!}=$ right hand
$endgroup$
– user62498
Dec 17 '18 at 7:37
add a comment |
$begingroup$
By hint Claude Leibovici,
$ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand
$e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$
Please let me know if this makes sense to you
$endgroup$
add a comment |
$begingroup$
By hint Claude Leibovici,
$ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand
$e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$
Please let me know if this makes sense to you
$endgroup$
add a comment |
$begingroup$
By hint Claude Leibovici,
$ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand
$e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$
Please let me know if this makes sense to you
$endgroup$
By hint Claude Leibovici,
$ sumlimits_{n=0}^{infty}sumlimits_{r=0}^{n}left(frac{1}{(n-r)!}a^{n-r}right)left(frac{1}{r!}b^{r}right)=sumlimits_{n=0}^{infty}frac{1}{n!}(a+b)^n=e^{a+b}$ on the other hand
$e^{a+b}=e^{a}e^{b}=left(sumlimits_{n=0}^{infty}frac{1}{n!}(a)^nright)left(sumlimits_{n=0}^{infty}frac{1}{n!}(b)^nright)$
Please let me know if this makes sense to you
answered Dec 17 '18 at 8:04
user62498user62498
1,948613
1,948613
add a comment |
add a comment |
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